Chemistry II

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If two or more chemical reactions can be combined to find an overall reactions, then their equilibrium constants can also be combined to find the equilibrium constant of the overall reaction. Given the following equilibrium reactions and constants, A + B ⇌ 2C K1 = 3.2 × 10-3 B + D ⇌ C K2 = 8.0 × 10-2 what is the equilibrium constant for the reaction A ⇌ B + 2D?

0.50 K1(1/K2)^2 = (3.2x10^-3)((1/8.0x10^-2)^2) = 0.50

A 32.8 mL sample of HCl is titrated with NaOH. If 42.7 mL of 0.425 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

0.553 M

Nuclear decay occurs according to first-order kinetics. Cobalt-60 decays with a rate constant of 0.131 years−1. After 5.00 years, a sample has a mass of 0.302 g. What was the original mass of the sample?

0.580 g

What is the pH of a 6.00 M H3PO4 solution? For H3PO4, Ka1 = 7.5 × 10−3, Ka2 = 6.2 × 10−8, and Ka3 = 4.2 × 10−13.

0.68

Give the complete symbol, with superscript and subscript, for a positron.

0/1e

A solution has a [H3O+] = 6.0 × 10−5 M at 25 °C. What is the [OH−] of the solution?

1.7 × 10⁻¹⁰ M

If ∆H = 348 kJ and ∆S = 164 J/K, the spontaneity of the reaction depends on temperature. Above what temperature in Celsius will the reaction be spontaneous?

1.85 × 103 °C

A solution has a pH of 9.73. What is the [H3O+] in the solution?

1.9 × 10−10 M

What is the pH of a 0.300 M solution of CH3NH2? Kb of CH3NH2 = 4.4 × 10−4

12.04 4.4x10^-4 = (X)(X)/(0.300) 1.3X10^-4 = X^2 X= 1.1x10^-2 pOH= -log[OH-] pOH= -log(1.1x10^-2) pOH= 1.96 pH + pOH = 14 14-1.96 = 12.04

What is the pH of a 0.280 M solution of (CH3)2NH? Kb of (CH3)2NH = 5.4 × 10−4

12.08

Given ∆Ssystem equals 345 J and ∆Ssurroundings equals −225 J, what is ∆Suniverse?

120. J

The isotope 122/53(I) decays by releasing a positron. What is the resulting isotope?121/53I->0/1e+?

121/52 Te

At 298 K, Kc = 18.3 for the following reaction. What is the value of Kp? C4(s) + 4 O2(g) ⇌ 4 CO2(g)

18.3 Kp= 18.3(0.0821 * 298) ^(4-4) Kp= 18.3

Based on the following reaction profile, how many intermediates are formed in the reaction A⟶D? (Figure 1) Express your answer as an integer.

What is the coefficient for OH−(aq) when MnO4−(aq) + H2S(g) → S(s) + MnO(s) is balanced in basic aqueous solution?

What is the overall reaction for the following half-reactions? Reduction half-reaction: Ag+(aq) + e− → Ag(s) Oxidation half-reaction: Ni(s) → Ni2+(aq)+ 2 e−

2 Ag+(aq) + Ni(s) → 2 Ag(s) + Ni2+(aq)

What is the pH of a solution that has 0.250 M HF and 0.250 M HClO? Ka of HF = 3.5 × 10−4 and Ka of HClO = 2.9 × 10−8

2.03 3.5x10^-4 = (X)(X) / (0.250) 8.8x10^-5 = X^2 X= 9.4x10^-3 pH= -log[H+] pH= -log(9.4x10^-3) pH= 2.03

What is the pH of a solution that has 0.200 M HF and 0.200 M HCN? Ka of HF = 3.5 × 10−4 and Ka of HCN = 4.9 × 10−10

2.08

What is the pOH for a solution with [OH−] = 8.3 × 10-3 M?

2.08

A 0.145 M solution of a weak acid has a pH of 2.75. What is the value of Ka for the acid?

2.2 × 10-5

What is the pH of a 0.200 M solution of HCOOH? Ka of HCOOH = 1.8 × 10−4

2.22 1.8x10^-4 = (X)(X)/(0.200) 3.6x10^-6 = X^2 X= 6.0x10^-3 pH= -log(6.0x10^-3) pH=2.22

A solution has a [H3O+] = 3.7 × 10−3 M. What is the pH of the solution?

2.43

A 0.200 M solution of a weak acid has a pH of 3.15. What is the value of Ka for the acid?

2.5 × 10-6

What is ΔG° for the following reaction at 25°C? CO(g) + 2 H2(g) ⇌ CH3OH(g) K = 2.1 × 10−4

21.0 kJ

What is ΔG° for the following reaction at 25°C? NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) K = 1.8 × 10−5

27.1 kJ

Give the symbol for an aluminum-27 nucleus using the

27/13Al

What is the change in the free energy, ∆G°, for the reaction? 6 CO2(g) + 6 H2O(g) → C6H12O6(s) + 6 O2(g) Substance ∆Gf° (kJ/mol) CO2(g) -394.4 H2O(g) -228.6 C6H12O6(g) -910.4

2827.0 kJ

A reaction has the general rate law, rate = k[A]2[B]. What is the order with respect to A and the overall order of the reaction?

2nd order with respect to A; 3rd order overall

How many transition states are there?

What is the coefficient for H2O(l) when SO32−(aq) + MnO4−(aq) → SO42−(aq) + Mn2+(aq) is balanced in acidic aqueous solution?

What is the coefficient for H2O(l) when Sn2+(aq) + IO3−(aq) → Sn4+(aq) + I−(aq) is balanced in acidic aqueous solution?

What is the overall reaction for the following half-reactions? Reduction half-reaction: Hg2+(aq) + 2 e− → Hg(l) Oxidation half-reaction: Cr(s) → Cr3+(aq) + 3 e−

3 Hg2+(aq) + 2 Cr(s) → 3 Hg(l) + 2 Cr3+(aq)

Nuclear decay occurs according to first-order kinetics. What mass of a 23 g chromium-51 sample is left after 82 days? The half-life of chromium-51 is 28 days.

3.0

The concentration of A is initially 0.100 M before it proceeds to equilibrium. What is the equilibrium concentration of B? A(g) ⇌ B(g) ΔG° = 31.5 kJ

3.05 × 10−7

A solution has a [H3O+] = 3.2 × 10−3 M at 25 °C. What is the [OH−] of the solution?

3.1 × 10⁻¹² M

A solution has a pH of 7.51. What is the [H3O+] in the solution?

3.1 × 10−8 M

If the half-life of a radioactive element is 15 hours, what percentage of the original sample would be left after 75 hours?

3.13%

What is the pOH for a solution with [OH−] = 4.6 × 10-4 M?

3.34

Nuclear decay occurs according to first-order kinetics. What mass of a 85.0 g iodine-123 sample is left after 18.7 hours? The half-life of iodine-123 is 13.2 hours.

31.8 g

What is the coefficient for H2O(l) when MnO4−(aq) + Fe2+(aq)→ Mn2+(aq) + Fe3+(aq) is balanced in acidic aqueous solution?

NO3−(aq) → NO(g) + 2 H2O(l) Select the half-reaction is properly balanced with respect to hydrogen. The half-reaction will not be properly balanced with respect to total charge.

4 H+(aq) + NO3−(aq)→ NO(g) + 2 H2O(l)

The equilibrium concentrations of A and B are 0.0434 M and 0.0923 M, respectively. What is ΔG° at 25°C? A(g) ⇌ 2 B(g)

4.04 kJ

What is ΔG° for the following reaction at 25°C? CoBr2(g) ⇌ CO(g) + Br2(g) Kc = 0.19

4.11 kJ

A reaction mixture initially has PN2 = 0.750 atm and PO2 = 0.750 atm. What is the equilibrium pressure of NO? N2(g) + O2(g) ⇌ 2 NO(g) Kp = 4.1 × 10-31 at 298 K

4.8 × 10-16 atm K= [NO]^2/ [N2][02] 4.1x10^-31 = (2X^2) / (0.750 - X)^2 square root of: 4.1x10^-31 = square root of: (2X^2) / (0.750)^2 6.4x10^-16 = 2X/0.750 4.8x10^-16 = 2X X= 2.4x10^-16 NO = 2X soo... 2(2.4x10^-16) = 4.8x10^-16 M

At 298 K, Kc = 1.2 × 10-3 for the following reaction. What is the value of Kp? CO(g) + Cl2(g) ⇌ COCl2(g)

4.9 × 10^-5

What is the pH of a 0.250 M solution of HCN? Ka of HCN = 4.9 × 10−10

4.96

The rate law for the following reaction is rate = k[H3PO4][I−][H+]2. What is the value of k based on the given data? H3PO4(aq) + 3 I−(aq) + 2 H+(aq) → H3PO3(aq) + I3−(aq) + H2O(l) Trial [H3PO4] (M) [I-] (M) [H+] (M) Initial rate (M/s) 1 0.150 0.150 0.150 0.220 2 0.300 0.150 0.150 0.440 3 0.150 0.450 0.150 0.660 4 0.300 0.150 0.300 1.760

435 M−3 s−1 rate = k[H3PO4][I−][H+]2 0.220 M/s = k(0.150)(0.150)(0.150)2 k = 435 M−3 s−1 rate = k[H3PO4][I−][H+]2 0.440 M/s = k(0.300)(0.150)(0.150)2 k = 435 M−3 s−1 rate = k[H3PO4][I−][H+]2 0.660 M/s = k(0.150)(0.450)(0.150)2 k = 435 M−3 s−1 rate = k[H3PO4][I−][H+]2 1.760 M/s = k(0.300)(0.150)(0.300)2 k = 435 M−3 s−1

A reaction mixture initially has [CO2] = 0.200 M and 10.0 grams of Ni. What is the equilibrium concentration of CO? Ni(s) + CO2(g) ⇌ NiO(s) + CO(g) K = 2.5 × 10-4 at 1500 K

5.0 × 10-5 M K= [CO] / [CO2] 2.5x10^-4 = (X) / (0.200) 5.0x10^-5 = X

What is the pH of a 0.150 M NH4Cl solution? Kb of NH3 = 1.76 × 10−5

5.035

What is the pH of a 0.225 M (CH3)2NH2Br solution? Kb of (CH3)2NH = 5.4 × 10−4

5.68

A solution has a pH of 3.24. What is the [H3O+] in the solution?

5.8 × 10−4 M

Nuclear decay occurs according to first-order kinetics. How long will it take for a sample of chromium-51 to decay from 35.0 grams to 10.0 grams? The half-life of chromium-51 is 28 days.

50. days

If the half-life of a radioactive element is 8 years, what percentage of the original sample would be left after 32 years?

6.25%

Nuclear decay occurs according to first-order kinetics. What mass of a 20.0 g lead-211 sample is left after 1.00 hour? The half-life of lead-211 is 36.1 minutes.

6.30 g

What is the pH of a neutral solution at a temperature when Kw = 2.2 × 10-14?

6.82

What is the pH of a neutral solution at a temperature when Kw = 1.8 × 10-14?

6.89

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− This rapid reaction gives the following rate data: [OCl−] (M) [I−] (M) Initial Rate (M/s) 1.5×10−3 1.5×10−3 1.36×10−4 3.0×10−3 1.5×10−3 2.72×10−4 1.5×10−3 3.0×10−3 2.72×10−4 Calculate the rate constant with proper units. Express the rate constant to two significant figures. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash (e.g. ft * lb or ft- lb).

60.4 M^-1*s^-1 k = rate / [OCl-] * [I-] k = 1.36*10^-4 M/s / [1.5 * 10^-3 M] * [1.5 * 10^-3 M]

Identify the balanced nuclear equation for the overall decay of 60/27Co resulting in beta and gamma emission after the two-step process.

60_27Co-> 60_28Ni+0_-1e+0_0y

What is the change in Gibbs free energy, ∆G, at 500°C given ∆H = −92.22 kJ and ∆S = −198.75 J/K? N2(g) + 3 H2(g) → 2 NH3(g)

61.41 kJ

What is the pH of a 0.100 M NaClO2 solution? Ka of HClO2 = 1.1 × 10−2

7.48

A reaction mixture initially has [CH3OH] = 0.500 M. What is the equilibrium concentration of H2? CH3OH(g) ⇌ CO(g) + 2 H2(g) K = 4.42 × 10-7 at 125 K

7.62 x 10-3 M K= [CO][H2]^2 / [CH3OH] 4.4x10^-7 = (X)(2X)^2 / (0.500) 2.21x10^-7 = 4X^3 X= 3.81x10^-3 H2= 2X sooo.... 2(3.81x10^-3) = 7.62 x 10-3 M

What is the coefficient for OH−(aq) when MnO4−(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq) is balanced in basic aqueous solution?

If 60 g of a radioactive substance naturally decays to 15 g after 16 hours, what is the half-life of the radioisotope?

8 hours

A solution has a [H3O+] = 8.4 × 10−9 M. What is the pH of the solution?

8.08

Nuclear decay occurs according to first-order kinetics. What is the half-life for mercury-206, which has a rate constant of 8.50 × 10-2 minutes-1?

8.15 minutes

What is ∆G when ∆G° is 2.80 kJ and Q equals 8.94 at 25°C? N2O4(g) ⇌ 2 NO2(g)

8.23 kJ

What is the pH of a 0.350 M MgF2 solution? Ka of HF = 3.5 × 10−4

8.65 (1.0x10^-14)/(3.5x10^-4)= 2.9x10^-11 square root of (2.9x10^-11)(0.700)= 4.5x10^-6 -log(4.5x10^-6)= 5.35 14-5.35 = 8.65

What is the pH of a 0.200 M KC7H5O2 solution? Ka of HC7H5O2 = 6.5 × 10−5

8.74

The half-life of chromium-51 is approximately 28 days. How long will it take for the radiation level of chromium-51 to drop to 12.5% (one-eighth) of its original level?

84 days

What is the pH of a 0.175 M solution of C5H5N? Kb of C5H5N = 1.7 × 10−9

9.23

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 5.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

9.56

A reaction mixture initially has [SO3] = 0.150 M. What is the equilibrium concentration of O2? 2 SO3(g) ⇌ 2 SO2(g) + O2(g) K = 1.6 × 10-10 at 350 K

9.7 × 10-5 M K= ([SO2]3[02]) / ([SO3]2 1.6x10^-10 = (2X)^2 (X)/ (0.150)^2 3.6x10^-12 = 4X^3 X= 9.7x10^-5 M

In the reaction shown below, which substance is oxidized? Al0(s) + Cr3+(aq) → Al3+(aq) + Cr0(s)

Al0(s)

What is the value of ∆G when a system is at equilibrium?

At equilibrium, ∆G = 0. At equilibrium, Q = K so ∆G = ∆G° + RT lnQ and ∆G° = −RT lnK which means ∆G must be equal to zero.

In the titration of a weak acid with a strong base with a 1:1 ratio, when does pH equal the pKa?

At half the volume it takes to reach the equivalence point

Which substance is the reducing agent in the following reaction? Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

CO(g)

Which reaction has a decrease in entropy?

CO(g) + 3 H2(g) → CH4(g) + H2O(g) has a decrease in entropy because the products have fewer moles of gas. H2O(s) → H2O(l) has an increase in entropy because the water goes from solid to liquid. 2 H2O(l) → 2 H2(g) + O2(g) has an increase in entropy because the products have more moles of gas and the reaction goes from liquid to gas. 2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) has an increase in entropy because there are more moles of products and the reaction goes from a solid and liquid to a solution and a gas.

Identify the formula for the conjugate base of HCO3−.

CO32−

In the reaction shown below, which substance is oxidized? Ca0(s) + Pb2+(aq) → Ca2+(aq) + Pb0(s)

Ca0(s)

Predict the type of particle emitted by carbon-14 and why.

Carbon-14 emits a beta particle because N/Z is too high. Carbon-14 has a N/Z ratio of 8/6, while the most common isotope, carbon-12, has an N/Z ratio of 6/6. Emitting a beta particle results in the formation of nitrogen-14, which has an N/Z ratio of 7/7, bringing the ratio closer to a stable value.

What is the molecularity of each of the following elementary reactions? Unimolecular/Bimolecular/Teremolecular? Cl2(g)->2Cl(g) NO(g) + Cl2(g) -> NOCl2(g) OCl- (aq) + H2O(l) ->HOCl(aq) + OH-(aq)

Cl2(g)->2Cl(g) - Unimolecular NO(g) + Cl2(g) -> NOCl2(g) - Bimolecular OCl- (aq) + H2O(l) ->HOCl(aq) + OH-(aq) - Bimolecular

Which substance is the reducing agent in the following reaction? Co(s) + Pt2+(aq) → Pt(s) + Co2+(aq)

Co(s) is the reducing agent as it causes Pt2+(aq) to be reduced to Pt(s). Pt2+(aq) is reduced as the oxidation state decreases from +2 to 0. Co(s) + Pt2+(aq) → Pt(s) + Co2+(aq) In the process, Co(s) is oxidized to Co2+(aq). The species that is oxidized is the reducing agent, and the species that is reduced is the oxidizing agent.

Analyzing a specific reaction Consider the following reaction: NO+O3→NO2+O2, rate=k[NO][O3] What would happen to the rate if [O3] were doubled?

Double

Chemical reactions and nuclear reactions have different kinetic behavior. What is the order for a nuclear decay reaction?

First order

When a hydrogen carbonate ion reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this

HCO₃⁻(aq) + H₂O(l) → H₃O+(aq) + CO₃²⁻(aq)

Select the strong acid. HNO2 HF HCl CH3COOH I DON'T KNOW YET

HCl

What is the equilibrium constant expression for the given reaction? CO2(g) + H2(g) ⇌ CO(g) + H2O(l)

K= [CO] / [CO2][H2]

Which of the following is the correct solubility product constant for the reaction shown below? Cu3(PO4)2(s) ⇌ 3 Cu2+(aq) + 2 PO43−(aq)

Ksp = [Cu2+]3[PO43−]2

Which of the following is the correct solubility product constant for the reaction shown below? SrCO3(s) ⇌ Sr2+(aq) + CO32−(aq)

Ksp = [Sr2+][CO32−]

What are the units for the rate constant, k, for a zero-order reaction?

M∙s−1

Identify the formula for the conjugate base of HPO42−.

PO43−

Consider the following equilibrium reaction and equilibrium constant: 2 SO2(g) + O2(g) ⇌ 2 SO3(g) K = 1.2 × 109 Is this equilibrium reaction a product favored equilibrium with mostly products, a reactant favored equilibrium with mostly reactants, or a true equilibrium with both reactants and products?

Product favored equilibrium with mostly products WHY? the equilibrium constant is a very large value. K= 1.2x10^9

In the reaction shown below, which substance is oxidized? 2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq)

Sn2+

What is the coefficient for OH−(aq) when Sn2+(aq) + IO3−(aq) → Sn4+(aq) + I−(aq) is balanced in basic aqueous solution?

The coefficient for OH−(aq) when Sn2+(aq) + IO3−(aq)→ Sn4+(aq) + I−(aq) is balanced in basic aqueous solution is 6. Write the two half-reactions for the equation. Sn2+(aq) → Sn4+(aq) IO3−(aq) → I−(aq) Balance the two half-reactions ((a) balance Sn and I, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen, (d) add electrons to balance charge). Sn2+(aq) → Sn4+(aq) + 2 e− IO3−(aq) + 6 H+(aq) + 6 e− → I−(aq) + 3 H2O(l) Multiply the half-reactions so the number of electrons is equal. 3 Sn2+(aq) → 3 Sn4+(aq) + 6 e− IO3−(aq) + 6 H+(aq) + 6 e− → I−(aq) + 3 H2O(l) Add the half-reactions together. 3 Sn2+(aq) + IO3−(aq) + 6 H+(aq) → 3 Sn4+(aq) + I−(aq) + 3 H2O(l) To balance in basic aqueous solution, look at the coefficient in front of H+ and add an equal number of OH− ions to each side of the equation. Then simplify so that H2O only appears on one side of the equation. 3 Sn2+(aq) + IO3−(aq) + 6 H+(aq) + 6 OH−(aq) → 3 Sn4+(aq) + I−(aq) + 3 H2O(l) + 6 OH−(aq) 3 Sn2+(aq) + IO3−(aq) + 6 H2O(l) → 3 Sn4+(aq) + I−(aq) + 3 H2O(l) + 6 OH−(aq) 3 Sn2+(aq) + IO3−(aq) + 3 H2O(l) → 3 Sn4+(aq) + I−(aq) + 6 OH−(aq)

Give the complete symbol, with superscript and subscript, for a gamma ray.

The complete symbol, with superscript and subscript, for a gamma ray is . A gamma ray is a form of high-energy electromagnetic waves. It has no mass or charge and is emitted as radiation.

A solution has a [Pb2+] of 0.00105 M. What concentration of chloride is needed before precipitation begins? Ksp of PbCl2 = 1.17 × 10-5

The concentration of chloride needed to begin precipitation is 0.106 M. The equation for the dissolution of PbCl2 is PbCl2(s) → Pb2+(aq) + 2 Cl−(aq), which gives the solubility constant expression Ksp = [Pb2+][Cl-]2 Substitute the given values and solve for [Cl−]. 1.17 × 10-5 = [0.00105][Cl-]2 1.11 × 10-2 = [Cl-]2 0.016 = [Cl-]

A solution has a [Ca2+] of 0.00350 M. What concentration of hydroxide is needed before precipitation begins? Ksp of Ca(OH)2 = 4.68 × 10-6

The concentration of hydroxide needed to begin precipitation is 0.0366 M. The equation for the dissolution of Ca(OH)2 is Ca(OH)2(s) → Ca2+(aq) + 2 OH−(aq), which gives the solubility constant expression Ksp = [Ca2+][OH−]2 Substitute the given values and solve for [OH−]. 4.68 × 10-6 = [0.00350][OH−]2 1.34 × 10-3 = [OH−]2 0.0366 = [OH−]

Which of the following is the correct solubility product constant for the following reaction? Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO43−(aq)

The correct solubility product constant for the reaction shown below is Ksp = [Ca2+]3[PO43−]2. Ca3(PO4)2(s) ⇌ 3 Cu2+(aq) + 2 PO43−(aq) According to the chemical equation, when Ca3(PO4)2(s) dissolves, the products are three calcium ions, 3 Ca2+(aq), and two phosphate ions, 2 PO43−(aq). Therefore, the concentration of Ca2+ needs to be raised to the third power, and the concentration of PO43- needs to be raised to the second power.

Which of the following is the correct solubility product constant for the following reaction? Zn(OH)2(s) ⇌ Zn2+(aq) + 2 OH−(aq)

The correct solubility product constant for the reaction shown below is Ksp = [Zn2+][OH−]2. Zn(OH)2(s) ⇌ Zn2+(aq) + 2 OH−(aq) According to the chemical equation, when Zn(OH)2(s) dissolves, the products are one zinc ion, Zn2+(aq), and two hydroxide ions, 2 OH−(aq). Therefore, the concentration of Zn2+ needs to be raised to the first power, and the concentration of OH- needs to be raised to the second power.

Which statement is true about the entropy for condensation?

The entropy decreases during condensation, so the change in entropy is a negative value. The entropy decreases during condensation, so the entropy of the products is less than the entropy of the reactants. Therefore, the change in entropy is negative.

Which statement is true about the entropy for vaporization?

The entropy increases during vaporization, so the change in entropy is a positive value. The entropy increases so the entropy of the products (gas) is greater than the entropy of the reactants (liquid). Therefore, the change in entropy will be positive.

The concentration of A is initially 0.200 M before it proceeds to equilibrium. What is the equilibrium concentration of B? A(g) ⇌ B(g) ΔG° = 17.5 kJ

The equilibrium concentration of B is 1.17 × 10−4 M. Use ΔG° and temperature to determine the equilibrium constant.

What is the equilibrium constant, K, for the following reaction at 25°C? 2 SO2(g) + O2(g) ⇌ 2 SO3(g) ΔG° = −148.6 kJ

The equilibrium constant is 1.14 × 1026. Use ΔG° and temperature to determine the equilibrium constant.

What is the equilibrium constant, K, for the following reaction at 25°C? HA(aq) → H+(aq) + A−(aq) ΔG° = −317 kJ

The equilibrium constant is 3.89 × 1055. Use ΔG° and temperature to determine the equilibrium constant.

Identify the half-life equation for substances that follow second-order kinetics.

The half-life equation for a second-order reaction is half-life equals 1 over the product of k and initial concentration of A. 1/K[ half-life equals initial concentration of A over 2k is the half-life equation for a zero-order reaction. half-life equals 0.693 over k is the half-life equation for a first-order reaction

What is the pH of a buffer system that has 0.11 M CH3COOH and 0.15 M CH3COO−? The pKa of acetic acid is 4.75.

The pH of a buffer system that has 0.11 M CH3COOH and 0.15 M CH3COO− is 4.88. Use the Henderson-Hasselbach equation to solve for the pH. The pKa value is given (4.75). The acid (HA) is CH3COOH and the conjugate base (A−) is CH3COO−. pH equals pKa plus log of the result of the concentration of A− divided by concentration of HA pH equals 4.75 plus log of the result of 0.15 divided by 0.11, which equals 4.75 plus 0.13 which is 4.88 The answer 4.62 is incorrect since this would result from putting the incorrect ratio (0.11/0.15) instead of (0.15/0.11) into the equation. The answer 6.11 is incorrect since this would result from not taking the log of the ratio. The answer 5.48 is incorrect since this would result from putting the incorrect ratio (0.11/0.15) instead of (0.15/0.11) into the equation and not taking the log of the ratio.

The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the concentration be after 12.0 minutes? Express your answer with the appropriate units. The integrated rate law for a first-order reaction is expressed as follows: [A]=[A]0e^−kt 0.1 * e^ - (4.5*10^-3 * 12 * 60)

For a second-order reaction, the rate constant k is the slope of the graph of 1[A] versus t. Based on this information and the data given, calculate the rate constant k for the reaction. Express your answer in M−1⋅min−1 to three significant figures.

The rate constant k = 3.10×10−2 M−1⋅min−1

Which reaction has an increase in entropy?

The reaction CaCO3(s) →CaO(s) + CO2(g) has an increase in entropy. The reaction begins with no gas and ends in the production of gas. Gases have a greater entropy than the solid. N2(g) + 3 H2(g) → 2 NH3(g) has a decrease in entropy because there are fewer moles of gas on the product side. H2O(g) → H2O(l) has a decrease in entropy because the phase changes from gas to liquid. 2 Mg(s) + O2(g) → 2 MgO(s) has a decrease in entropy because there are fewer moles of gas on the product side.

The equilibrium constant for the given reaction is 2.0 at 473 K. After 10 seconds, [COF2] = 0.050 M, [CO2] = 0.15 M, and [CF4] = 0.15 M. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? 2 COF2(g) ⇌ CO2(g) + CF4(g)

The reaction will proceed to the left because Q > K. Q= [CO2][CF4] / [COF2]2 Q=[0.15][0.15] /[0.050]2 Q= 9.0 Q is greater than K, so the reaction will proceed to the left. As the concentrations of the products decrease and the concentration of the reactant increases, the value of Q will decrease until it reaches the value of K.

The equilibrium constant for the given reaction is 0.0142 at 298 K. After 20 seconds, [NOBr] = 0.15 M, [NO] = 0.020 M, and [Br2] = 0.010. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? 2 NOBr(g) ⇌ 2 NO(g) + Br2(g)

The reaction will proceed to the right because Q < K. Q= [NO]2[Br2] / [NONr]2 Q= [0.020]2[0.010] / [0.15]2 Q= 1.8 x10^-4 so... Q is less than K, so the reaction will proceed to the right. As the concentrations of the products increase and the concentration of the reactant decreases, the value of Q will increase until it reaches the value of K.

The titration of NH3 with HCl produces a salt and water. The resulting salt is _________.

The titration of NH3 with HCl produces a salt and water. The resulting salt is acidic because NH3 is a weak base and HCl is a strong acid. When a weak base reacts with a strong acid, the resulting salt contains the cation of the weak base and the anion of the strong acid. The anion will not undergo hydrolysis but the cation will. When the cation undergoes hydrolysis, it will produce H3O+ ions in solution. NH4+ + H2O ⇌ NH3 + H3O+

How many electrons are needed to balance each of the following half-reactions? Reduction half-reaction: Fe3+(aq) + _e− → Fe(s) Oxidation half-reaction: Al(s) → Al3+(aq)+ _e− Overall reaction: Fe3+(aq) + Al(s) → Fe(s) + Al3+(aq)

Three electrons are needed to balance each of the half-reactions. Reduction half-reaction: Fe3+(aq) + 3 e− → Fe(s) Oxidation half-reaction: Al(s) → Al3+(aq) + 3 e− Overall reaction: Fe3+(aq) + Al(s) → Fe(s) + Al3+(aq) Since same number of electrons are present on the left side of the first half-reaction and the right side of the second half-reaction, they cancel out. Notice that both sides of the overall redox reaction have that the same total charge and the same number and types of atoms.

A zero-order reaction has a constant rate of 3.40×10−4 M/s. If after 75.0 seconds the concentration has dropped to 3.50×10−2 M, what was the initial concentration? Express your answer with the appropriate units.

[A]0[A]0 = 6.05×10−2 MM [A] - [Ao] = kt [Ao] = 3.50*10^-2 + 3.40*10^-4 M/s * 75 s

For the titration of a weak base with a strong acid, the pH at the half-equivalence point volume is _________.

basic

Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k= 0.011 day−1. Which one decays at a faster rate?

iodine-125

Nuclear decay follows first-order kinetics. Which graph will result in the formation of a straight line?

ln[A] vs. time

Which equation shows the integrated rate law for a substance that reacts according to first-order kinetics?

ln[A]t = -kt + ln[A]0

Nuclear decay occurs according to first-order kinetics. What equations give the correct integrated rate law for a first-order process?

ln[At] = -kt + ln[Ao]

For the reaction A⟶D, is ΔE positive, negative, or zero?

positive

The reaction 2ClO2(aq)+2OH−(aq)→ClO3−(aq)+ClO2−(aq)+H2O(l) was studied and the results listed in the table were obtained. Experiment [ClO2](M) [OH−](M) Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 Use the rate constant you determined in Part B to calculate the rate of reaction when [ClO2]= 0.150 M and [OH−]= 5.0×10−2 M . Express the rate in molarity per second to two significant digits.

rate = 0.26 M/s The rate is calculated according to the rate law determined in Part A. Multiplying k , [ClO2]2 , and [OH−] will yield the rate of reaction under the given conditions.

The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2 The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82×10−3s−1. What is the rate of reaction when [N2O5]=0.0240 ? Express the rate in molarity per second to three significant figures.

rate = 1.16×10−4 M/s The rate law in Part A is used to determine the rate of the reaction at any concentration of the reactant. rate=4.82×10−3 s−1[N2O5] =4.82×10−3 s−1(0.0240 M) =1.16×10−4 M/s

Which expression best defines the rate of the reaction with respect to CO2, given the following reaction? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

rate = 1/3 - (▵[CO2] / ▵t) The equation that best represents the rate of change of O2 for the following reaction is rate equals 1 over 3 times change in concentration of carbon dioxide over change in time. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Carbon dioxide is a product which is being formed so the equation must have a positive sign. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is ⅓ of the change in concentration of carbon dioxide over the change in time.

Which expression best defines the rate of the reaction with respect to H2O, given the following reaction? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

rate = 1/4 - ▵[H2O] / ▵t The equation that best represents the rate of change of H2O for the following reaction is rate equals 1 over 4 times change in concentration of water over change in time. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Water is a product which is being formed so the equation must have a positive sign. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is ¼ of the change in concentration of water over the change in time.

The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2 The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82×10−3s−1. What is the rate when the concentration of N2O5 from Part B is doubled? [N2O5 = 0.0240] Express the rate in molarity per second to three significant figures.

rate = 2.31×10−4 M/s Since the reaction is first order with respect to the reactant concentration, doubling [N2O5] doubles the rate.

The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2 The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82×10−3s−1. [N2O5 = 0.0240] What is the rate when the concentration of N2O5 from Part B is halved? Express the rate in molarity per second to three significant figures.

rate = 5.78×10−5 M/s

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically. The reaction of compound A forming compound B was studied and the following data were collected: Time (s) [A] (M) 0. 0.184 200. 0.129 500. 0.069 800. 0.031 1200. 0.019 1500. 0.016 What is the instantaneous rate of the reaction at t=800. s? Express your answer to two significant figures and include the appropriate units.

rate = 6.8×10−5 Ms The instantaneous rate can also be determined by using calculus. The slope of the tangent line is the first derivative of the line of the graph expressed as d[A]/dt versus the designation used for the average rate, Δ[A]/Δt

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− This rapid reaction gives the following rate data: [OCl−] (M) [I−] (M) Initial Rate (M/s) 1.5×10−3 1.5×10−3 1.36×10−4 3.0×10−3 1.5×10−3 2.72×10−4 1.5×10−3 3.0×10−3 2.72×10−4 Calculate the rate when [OCl−]=2.2×10−3 and [I−]=5.8×10−4 . Express the rate in molarity per second to two significant figures.

rate = 7.7×10−5 M/s k = rate / [OCl-] * [I-] 60 M^-1*s^-1 = rate / [2.2×10−3] * [5.8*10-4]

What are the units for the rate constant, k, for a first-order reaction?

s−1

Identify the half-life equation for substances that follow first-order kinetics.

t(1/2) = 0.693/k

t1/2 = 430 yr t1/2 = 0.693/k

The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 56 torr/min , what is the rate of change of the total pressure of the vessel?

ΔPtotal/Δt = 28 torr/min d[NO]/dt = 56 torr/min d[Cl2]/dt = 28 torr/min d[NO]/dt = 56 torr/min Rate change = Reactant change - Product change (56+28) - 56 28 torr/min

Consider the combustion of H2(g):2H2(g)+O2(g)→2H2O(g). If hydrogen is burning at the rate of 0.48 mol/s , what is the rate of consumption of oxygen?

Δ[O2]/Δt = 0.24 mol/s

What is the rate of change of C3H8 if the rate of change of O2 is −0.300 M/s? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

−0.0600 M/s

What is the oxidation state of sulfur in Na2S2O3?

What is the oxidation state of boron in Na2B4O7?

What is the oxidation state of chlorine in HClO2?

What is the oxidation state of nitrogen in Mg(NO2)2?

What is the oxidation state of silicon in SiO32−?

What is the oxidation state of iodine in IO3−?

What is the oxidation state of nitrogen in N2O5?

What is the oxidation state of phosphorus in H3PO4?

What is the oxidation state of sulfur in H2SO4?

What is the oxidation state of manganese in KMnO4?

What is ∆G when ∆G° is −1010. kJ and Q equals 18.97 at 25°C? 4 NH3(g)+ 5 O2(g) ⇌ 4 NO(g)+ 6 H2O(g)

-1003 kJ

Which expression is true for the given reaction? N2O4(g) ⇌ 2 NO2(g)

-▵[N2O4]/ ▵t = 1/2 -(▵[NO2]/ ▵t) The true expression for the reaction N2O4(g) ⇌ 2 NO2(g) is negative change in concentration of N2O4 over change in time equals one-half times the change in concentration of NO2 over change in time. N2O4 is a reactant and is being consumed so a negative signed is needed. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is equal to the change in concentration of N2O4 over time, therefore rate equals negative change in concentration of N2O4 over change in time. NO2 is a product and is being formed so a positive sign is needed for its term. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is ½ of the change in concentration of NO2 over the change in time, therefore rate equals one-half times the change in concentration of NO2 over change in time. Since negative change in concentration of N2O4 over change in time and one-half times change in concentration of NO2 over change in time are both equal to rate, they must be equal to each other.

What is the oxidation state of carbon in glucose, C6H12O6?

Now consider the following reaction and data: H2+2IBr→2HBr+I2 Time(s) I2 concentration (M) 5 1.36 15 1.734 Based on your answer to Part B or C, what is the average rate of change of H2? Remember that reactant concentrations decrease over time. Express your answer to three decimal places and include the appropriate units.

0.037 M/s (1.734 - 1.36) / (15-5)

Now consider the following reaction and data: H2+2IBr→2HBr+I2 Time(s) I2 concentration (M) 5 1.36 15 1.734 What is the average rate of formation of I2? Express your answer to three decimal places and include the appropriate units.

0.037 M/s (1.734 - 1.36) / (15-5)

Now consider the following reaction and data: H2+2IBr→2HBr+I2 Time(s) I2 concentration (M) 5 1.36 15 1.734 Based on your answer to Part B, what is the average rate of formation of HBr?

0.074 M/s Double the rate of formation of I2

Nuclear decay occurs according to first-order kinetics. A nuclide decays in 23.0 minutes from 12.9 g to 2.04 g. What is the rate constant for the nuclide?

0.0803 min^-1

The concentration of A is initially 0.150 M before it proceeds to equilibrium. What is the equilibrium concentration of B? A(g) ⇌ B(g) ΔG° = −2.25 kJ

0.107 M

A 54.6 mL sample of HCl is titrated with NaOH. If 26.5 mL of 0.255 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution? f NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

0.124 M

0.13 mg

A 14.6 mL sample of H3PO4 is titrated with NaOH. If 9.32 mL of 0.927 M NaOH is needed to reach the endpoint, what is the concentration (M) of the H3PO4 solution? 3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

0.197 M

A 25.0 mL sample of HCl is titrated with NaOH. If 28.9 mL of 0.175 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

0.202 M

A 19.5 mL sample of Ba(OH)2 is titrated with HCl. If 28.4 mL of 0.394 M HCl is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l)

0.287 M

A 25.0 mL sample of H3PO4 is titrated with NaOH. If 29.2 mL of 0.738 M NaOH is needed to reach the endpoint, what is the concentration (M) of the H3PO4 solution? 3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

0.287 M

A substance, A, decomposes according to zero-order kinetics with a rate constant, k = 3.80 × 10−2 M∙s−1, that decays for 24.5 seconds from its initial concentration of 1.25 M. What is the final concentration of A?

0.319 M [A]t = -kt + [A]0 [A]t = -(3.80×10-2 M·s-1)(24.5 s) + 1.25 M [A]t = 0.319 M The final concentration of A is 0.319 M. Use the integrated rate law for a zero-order reaction with the given values to determine the final concentration. [A]t = -kt + [A]0 [A]t = -(3.80×10-2 M·s-1)(24.5 s) + 1.25 M [A]t = 0.319 M 0.493 M is the concentration if you used the equation for the integrated rate law equation for a first-order reaction. 0.578 M is the concentration if you used the equation for the integrated rate law equation for a second-order reaction.

At 298 K, Kc = 0.0142 for the following reaction. What is the value of Kp? 2 NOBr(g) ⇌ 2 NO(g) + Br2(g)

0.347 Kp= Kc(RT)^change in "n" Kp= 0.0142 (0.0821 * 298) ^(3-2) Kp= 0.347

If two or more chemical reactions can be combined to find an overall reactions, then their equilibrium constants can also be combined to find the equilibrium constant of the overall reaction. Given the following equilibrium reactions and constants, A + C ⇌ B + D K1 = 1.5 C ⇌ B K2 = 3.9 what is the equilibrium constant for the reaction A ⇌ D?

0.38

Nuclear decay occurs according to first-order kinetics. A nuclide decays in 3.40 days from 45.0 g to 12.1 g. What is the rate constant for the nuclide?

0.388 days^-1

A 7.61 mL sample of Ba(OH)2 is titrated with H3PO4. If 17.4 mL of 0.139 M H3PO4 is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? 3 Ba(OH)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(aq) + 6 H2O(l)

0.477 M

Consider the combustion of H2(g):2H2(g)+O2(g)→2H2O(g). If hydrogen is burning at the rate of 0.48 mol/s , What is the rate of formation of water vapor?

0.48 mol/s

A 32.9 mL sample of Ba(OH)2 is titrated with HCl. If 37.2 mL of 0.863 M HCl is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l)

0.488 M

Nuclear decay occurs according to first-order kinetics. What is the half-life of zinc-65 if a sample decays from 65.0 g to 2.90 g in 3.0 years?

0.69 years

What mass of NaCN must be added to create 250.0 mL of a buffer solution with a pH = 9.00 and an HCN concentration of 0.100 M? The pKa of hydrocyanic acid is 9.23.

0.72 g The mass of NaCN needed is 0.72 g. Use the Henderson-Hasselbalch equation to solve for the concentration of CN− needed. pH equals pKa plus log of the result of the concentration of A minus divided by the concentration of HA pH equals pKa plus log of the result of the concentration of CN minus divided by the concentration of HCN 9.00 equals 9.23 plus log of the result of the concentration of CN minus divided by 0.100. So -0.23 equals log of the result of the concentration of CN minus divided by 0.100. Undo the log to find that 10 to the power of negative 0.23 equals concentration of CN minus divided by 0.100. So the concentration of CN minus equals 0.059 Use the concentration of CN−, the volume of the solution, and the molar mass of NaCN to find the mass of NaCN. 250.0 milliliters times 1 liter over 1000 milliliters times 0.059 moles CN minus over liter times 1 mole NaCN over 1 mole CN minus times 49.01 grams over 1 mole NaCN equals 0.72 grams NaCN The answer 2.9 g is incorrect because it is based on a volume of 1 L rather than 250. mL. The answer 7.2 g is incorrect because it uses [CN−] = 0.59 M instead of 0.059 M. The answer 2.1 g is incorrect because it switches the pH and the pKa in the calculation of [CN−].

What is the rate of change of H2O if the rate of change of CO2 is 0.600 M/s? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

0.800 M/s If the rate of change of CO2 is 0.600 M/s, then the rate of change of H2O is 0.800 M/s. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) The equation that compares the rate of change of CO2 and H2O is the following because both terms are equal to rate. negative 1 over 4 times change in concentration of H2O over change in time equals 1 over 3 times change in concentration of CO2 over change in time Insert the given rate of change for CO2 to solve for the rate of change of H2O.

A 28.8 mL sample of HCl is titrated with NaOH. If 39.5 mL of 0.639 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

0.876 M

A 18.5 mL sample of Ba(OH)2 is titrated with HCl. If 37.5 mL of 0.921 M HCl is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l)

0.933 M

A 16.1 mL sample of Ba(OH)2 is titrated with H3PO4. If 32.7 mL of 0.314 M H3PO4 is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? 3 Ba(OH)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(aq) + 6 H2O(l)

0.957 M

Give the symbol to complete the balanced nuclear equation and describe the type of radiation.66/29 Cu-> ? + 66/30 Zn

0/-1e, beta decay

Give the symbol to complete the balanced nuclear equation.11/6C-> ?+11/5B

0/1e

What is the coefficient for H2O(l) when SO42−(aq) + Br2(l) → S2O32−(aq) + BrO3−(aq) is balanced in acidic aqueous solution?

A 22.9 mL sample of NaOH is titrated with H3PO4. If 17.9 mL of 0.427 M H3PO4 is needed to reach the endpoint, what is the concentration (M) of the NaOH solution? 3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

1.00 M

1.00 mg

A reaction mixture initially has [Br2] = 0.100 M and [Cl2] = 0.100. What is the equilibrium concentration of BrCl? Br2(g) + Cl2(g) ⇌ 2 BrCl(g) K = 1.45 × 10-4 at 150 K

1.19 × 10-3 M K= [BrCl]^2 / [Br2][Cl2] 1.45x10^-4 = (2X)^2 / (0.100) square root of: 1.45x10^-4 = square root of: (2X)^2 / (0.100) 1.20x10^-2 = (2X)/0.100 1.20x10^-3 = 2.01X X= 5.97x10^-4 [BrCl] = 2X soo... 2(5.97x10^-4) = 1.19 × 10-3 M

What is the molar solubility of Ca(OH)2 if the Ksp is 6.5 × 10-6? Ca(OH)2(s) ⇌ Ca2+(aq) + 2 OH−(aq)

1.2 × 10-2 M The molar solubility of Ca(OH)2 is 1.2 × 10-2 M. Ca(OH)2(s) ⇌ Ca2+(aq) + 2 OH−(aq) Ksp = [Ca2+][OH−]2 [x][2x]2 = 6.5 × 10-6 4x3 = 6.5 × 10-6 x = 1.2 × 10-2 The [Ca2+] is equal to the amount of Ca(OH)2 that dissolves, which is the molar solubility.

Nuclear decay occurs according to first-order kinetics. What is the rate constant for carbon-14, which has a half-life of 5730 years?

1.21 x 10^-4 years^-1

Nuclear decay occurs according to first-order kinetics. What is the rate constant for radon-220, which has a half-life of 55.6 seconds?

1.25 × 10^-2 sec^-1

If ∆H = 498 kJ and ∆S = 319 J/K, the spontaneity of the reaction depends on temperature. Above what temperature will the reaction be spontaneous?

1.29 × 103 °C

By what factor does the rate constant increase when the temperature increases from 200. K to 400. K for a reaction with an activation energy of 845 J/mol?

1.3

A solution has a [H3O+] = 7.9 × 10−11 M at 25 °C. What is the [OH−] of the solution?

1.3 × 10⁻⁴ M

A solution has a pH of 5.87. What is the [H3O+] in the solution?

1.3 × 10−6 M

What is the activation energy for a reaction when the rate constant doubles when the temperature increase from 150. K to 450. K?

1.30 × 10^3 J/mol ln 2 = (Ea/ 8.314) ((1/150) - (1/450)) Ea = 1.30 x10^3 J/ mol The activation energy is 1.30 × 103 J/mol. 563 J/mol is the activation energy if log is used instead of the natural log. 18.8 J/mol is the activation energy if the ideal gas constant is in the numerator of the equation instead of the denominator.

What is the half-life of a first-order reaction with a rate constant of 1.00×10−4 s−1? Express your answer with the appropriate units.

1.34×10−3 s−1 The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A]=[A]0e−kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693/k This equation calculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life.

What is the rate constant of a first-order reaction that takes 517 seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units

1.34×10−3 s−1 Use t1/2 = 0.693/k

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 25.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

1.353 The pH after 25.00 mL of HCl has been added is 1.353. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCl + NH3 → NH4Cl + H2O I 5.00 × 10−3 mol 3.00 × 10−3 mol 0 ~ C −3.00 × 10−3 mol −3.00 × 10−3 mol +3.00 × 10−3 mol ~ F 2.00 × 10−3 mol 0 3.00 × 10−3 mol ~ After the NH3 and HCl react, 2.00 × 10−3 moles of HCl and 3.00 × 10−3 moles of NH4+ remain in a total volume of 45.00 mL (25.00 mL HCl + 20.00 mL NH3). Since HCl is a strong acid, the presence of NH4+ does not contribute to the pH. The [HCl] is calculated using the moles of HCl and the total volume of solution. concentration of HCl equals moles of HCl divided by volume of solution. concentration of HCl equals 2.00 × 10^-3 moles HCl divided by 0.0450 liters, so concentration of HCl equals 4.44 × 10^-2 molar. HCl fully dissociates so [HCl] = [H+]. pH = −log(4.44 × 10−2) pH = 1.353 12.647 is incorrect because it is the pOH, not the pH. 1.125 is incorrect because it uses the molarity of NH4+ instead of the molarity of HCl. 2.000 is incorrect because it uses the moles of HCl instead of the molarity of HCl.

A 25.00 mL sample of 0.175 M HCl is being titrated with 0.250 M NaOH. What is the pH after 11.50 mL of NaOH has been added?

1.386

A reaction mixture initially has [H2S] = 0.100 M. What is the equilibrium concentration of H2? 2 H2S(g) ⇌ 2 H2(g) + S2(g) K = 1.67 × 10-7 at 298 K

1.49 × 10-3 M 1.67x10^-7 = (2X)^2(X) / (0.100)^2 1.67x10^-7 = 4X^3 X= 7.47x10^-4 H2= 2X so.... 2(7.47x10^-4) = 1.49x10^-3

A 12.1 mL sample of Ba(OH)2 is titrated with H3PO4. If 35.4 mL of 0.351 M H3PO4 is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? 3 Ba(OH)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(aq) + 6 H2O(l)

1.54 M

Rates of nuclear decay are measured in units of half-life (t1/2). The half-life of a radioisotope is the amount of time it takes for one-half of the sample to decay. If the half-life of a radioactive element is 18 days, what percentage of the original sample would be left after 108 days?

1.56%

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 20.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

1.602

What is the activation energy for a reaction when the rate constant increases by a factor of 3.5 when the temperature increase from 125 K to 625 K?

1.63 × 10^3 J/mol ln 3.5 = Ea/ 8.314 (1/125 - 1/625) Ea = 1.63 x 10^3 J/mol The activation energy is 1.63 × 103 J/mol. natural log of k2 over k1 equals activation energy over R times the result of 1 over T1 minus 1 over T2 natural log of 3.5 equals activation energy over 8.314 Joules per mole Kelvin times result of 1 over 125 Kelvin minus 1 over 625 Kelvin, so the activation energy equals 1.63 times 10^3 Joules per mole 707 J/mol is the activation energy if log is used instead of the natural log. 23.5 J/mol is the activation energy if the ideal gas constant is in the numerator of the equation instead of the denominator.

A substance, A, decomposes according to zero-order kinetics from an initial concentration of 1.90 M to 0.420 M in 85.2 seconds. What is the value of rate constant, k?

1.74 × 10−2 M∙s−1 The rate constant for the reaction is 1.74 × 10-2 M∙s−1. Use the integrated rate law for a zero-order reaction with the given values to determine the rate constant. [A]t = -kt + [A]0 0.420 M = -k(85.2 s) + 1.90 M k = 1.74 × 10-2 M·s-1 1.78 × 10−2 M∙s−1 is the rate constant if you used the equation for the integrated rate law equation for a first-order reaction. 2.18 × 10−2 M∙s−1 is the rate constant if you used the equation for the integrated rate law equation for a second-order reaction.

A 0.110 M solution of a weak acid has a pH of 2.84. What is the value of Ka for the acid?

1.8 × 10-5

A reaction mixture initially has [SO3] = 0.150 M. What is the equilibrium concentration of SO2? 2 SO3(g) ⇌ 2 SO2(g) + O2(g) K = 1.6 × 10-10 at 350 K

1.9 × 10-4 M K=[SO2]^2[02] / [SO3]^2 1.6x10^-10 = (2X)^2(X) / (0.150)^2 3.6X10^-12 = 4X^3 X= 9.7x10^-5 SO2 = 2X soo.... 2(9.7x10^-5) = 1.9 × 10-4 M

What is the pH of a solution that has 0.300 M HNO2 and 0.300 M HCN? Ka of HNO2 = 4.6 × 10−4 and Ka of HCN = 4.9 × 10−10

1.92

What is the pH of a 0.375 M solution of HF? Ka of HF = 3.5 × 10−4

1.96

Which equation shows the integrated rate law for a substance that reacts according to second-order kinetics?

1/[A]t = kt + 1/[A]0

A solution has a [H3O+] = 7.9 × 10−11 M. What is the pH of the solution?

10.10

What is the pOH for a solution with [H3O+] = 3.2 × 10-4 M?

10.51

What is the pH of a buffer system that has 0.35 M CH3NH2 and 0.42 M CH3NH3+? The Kb of methylamine is 4.4 × 10−4.

10.56

The rate law for the following reaction is rate = k[F2][ClO2]. What is the value of k based on the given data? F2(g) + 2 ClO2(g) → 2 FClO2(g) Trial [F2] (M) [ClO2] (M) Initial rate (M/s) 1 0.050 0.025 0.130 2 0.050 0.050 0.260 3 0.200 0.025 0.520

104 M−1 s−1 rate = k[F2][ClO2] 0.130 M/s = k(0.050)(0.025) k = 104 M−1 s−1 rate = k[F2][ClO2] 0.260 M/s = k(0.050)(0.050) k = 104 M−1 s−1 rate = k[F2][ClO2] 0.520 M/s = k(0.200)(0.025) k = 104 M−1 s−1

Give the symbol for a silver-108 nucleus using the

108/47Ag

What is the pH of a 0.135 M NaCN solution? Ka of HCN = 4.9 × 10−10

11.20 why? 2.0 x 10^-5 = (X)(X)/ (0.135) 2.7 x 10^-6 = X^2 X= 1.6 x 10^-3 pOH= -log[OH-] pOH= -log[1.6x10^-3) pOH= 2.80 pH + pOH= 14 so.... 14-2.80 = 11.20

What is the pH of a 0.150 M solution of NH3? Kb of NH3 = 1.76 × 10−5

11.210

What is the pH of a 0.145 M solution of (CH3)3N? Kb of (CH3)3N = 6.4 × 10−5

11.48

What is the coefficient for H2O(l) when NH3(g) + NO2(g) → N2(g) is balanced in acidic aqueous solution?

A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 30.0 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10−4

12.398

Nuclear decay occurs according to first-order kinetics. What is the half-life of europium-152 if a sample decays from 10.0 g to 0.768 g in 50.0 years?

13.5 years

Nuclear decay occurs according to first-order kinetics. Argon-37 decays with a rate constant of 0.0198 days−1. After 98.3 days, a sample has a mass of 2.14 g. What was the original mass of the sample?

15.0 g

Identify the balanced nuclear equation for the beta decay of 157/63 Eu.

157/63Eu-> 0/-1e+157/64Gd

Given the rate law, rate = k[A]2[B]2, the rate will increase by a factor of ____ when the concentration of A and B are both doubled.

16 Given the rate law, rate = k[A]2[B]2, the rate will increase by a factor of 16 when the concentration of A and B are both doubled. If the initial [A] is x, then the final [A] is 2x. Therefore, the rate will increase by a factor of 22, which is 4. If the initial [B] is y, then the final [B] is 2y. Therefore, the rate will increase by a factor of 22, which is 4. The total change in rate will increase by a factor of 4 × 4 = 16.

Given the rate law, rate = k[A][B]3, the rate will increase by a factor of ____ when the concentration of A and B are both doubled.

16 Given the rate law, rate = k[A][B]3, the rate will increase by a factor of 16 when the concentration of A and B are both doubled. If the initial [A] is x, then the final [A] is 2x. Therefore, the rate will increase by a factor of 21, which is 2. If the initial [B] is y, then the final [B] is 2y. Therefore, the rate will increase by a factor of 23, which is 8. The total change in rate will increase by a factor of 2 × 8 = 16.

Identify the balanced nuclear equation for the positron emission of 18/9F.

18/9F->0/1e+18/8O

What is the standard entropy change for the reaction? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Substance ∆S° (J/mol∙K) NH3(g) 193 O2(g) 205 NO(g) 211 H2O(g) 189

181 J/K

The isotope ^192_78Pt decays by releasing an alpha particle. What is the resulting isotope?192_78Pt->He+?

188_76Os

A substance reacts with zero-order kinetics and its concentration goes from 1.30 M to 0.89 M in 12 minutes. What is its half-life?

19 min [A]t = -kt + [A]0 0.89 M = -k(12 min) + 1.30 M k = 3.4 × 10-2 M·min-1 t(1/2) = [1.30M]o / 2(3.4 x 10^-2) t(1/2) = 19 min The half-life is 19 minutes. Use the integrated rate law for a zero-order reaction with the given values to determine the rate constant. [A]t = -kt + [A]0 0.89 M = -k(12 min) + 1.30 M k = 3.4 × 10-2 M·min-1 Then use the half-life equation to find the half-life, t1/2. half-life equals concentration of the initial concentration of A divided by 2k. half-life equals 1.30 molar divided by the product of 2 times 3.4 times 10^−2 molar per minute. half-life equals 19 minutes 20 min is the half-life if you used the equation for the half-life equation for a first-order reaction. 23 min is the half-life if you used the equation for the integrated rate law equation for a second-order reaction.

A reaction has the general rate law, rate = k[A][B]. What is the order with respect to A and the overall order of the reaction?

1st order with respect to A; 2nd order overall For a reaction with the general rate law, rate = k[A][B], the reaction is 1st order with respect to A and 2nd order overall. The order with respect to a particular substance is equal to its superscript in the rate law. The overall order for a reaction is the sum of the orders with respect to each reactant.

A reaction has the general rate law, rate = k[A][B][C]. What is the order with respect to A and the overall order of the reaction?

1st order with respect to A; 3rd order overall For a reaction with the general rate law, rate = k[A][B][C], the reaction is 1st order with respect to A and 3rd order overall. The order with respect to a particular substance is equal to its superscript in the rate law. The overall order for a reaction is the sum of the orders with respect to each reactant.

A reaction has the general rate law, rate = k[A][B]2[C]. What is the order with respect to A and the overall order of the reaction?

1st order with respect to A; 4th order overall For a reaction with the general rate law, rate = k[A][B]2[C], the reaction is 1st order with respect to A and 4th order overall. The order with respect to a particular substance is equal to its superscript in the rate law. The overall order for a reaction is the sum of the orders with respect to each reactant.

Analyzing a specific reaction Consider the following reaction: NO+O3→NO2+O2, rate=k[NO][O3] What is the overall reaction order?

2 Learning Goal: To understand reaction order and rate constants. For the general equation aA+bB→cC+dD, the rate law is expressed as follows: rate=k[A]m[B]n where m and n indicate the order of the reaction with respect to each reactant and must be determined experimentally and k is the rate constant, which is specific to each reaction. Order For a particular reaction, aA+bB+cC→dD, the rate law was experimentally determined to be rate=k[A]0[B]1[C]2=k[B][C]2 This equation is zero order with respect to A. Therefore, changing the concentration of A has no effect on the rate because [A]0 will always equal 1. This equation is first order with respect to B. This means that if the concentration of B is doubled, the rate will double. If [B] is reduced by half, the rate will be halved. If [B] is tripled, the rate will triple, and so on. This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9, and so on.

HI decomposes into its elements according to second-order kinetics. How long will it take for the concentration to decrease to 1.25 M from an initial concentration of 2.25 M? The rate constant, k, equals 1.6 × 10−3 M−1hr−1. 2 HI(g) → H2(g) + I2(g)

2.2 × 10^2 hours 1/[1.25]t = (1.6x10^-3)t + 1/[2.25M]o 0.356 = (1.6x10^-3)t t= 2.2x10^2 hr he time to go from 2.25 M to 1.25 M is 2.2 × 102 hour. Use the integrated rate law for second-order reactions with the given values to find the time. one over the concentration of A at time t equals kt plus one over the initial concentration of A. one over 1.25 equals 1.6 times 10^−3 inverse molar times inverse hours times time plus 1 over 2.25 molar. 3.56 equals 1.6 times 10^−3 inverse molar times inverse hours times time. time equals 2.2 times 10^2 hours 6.3 × 102 hours is the time using the zero-order integrated rate law. 3.7 × 102 hours is the time using the first-order integrated rate law.

HI decomposes into its elements according to second-order kinetics. What concentration of HI remains from an initial concentration of 2.30 M after 4.5 hours? The rate constant, k, equals 1.6 × 10−3 M−1hr−1. 2 HI(g) → H2(g) + I2(g)

2.26 M 1/[A]t = (1.6 x10^-3)(4.5hr) + 1/[2.30M]0 1/[A]t = 0.442 [A]t = 2.26 M The concentration after 4.5 hours is 2.26 M. Use the integrated rate law for second-order reactions with the given values to find the final concentration. 2.29 M is the concentration using the zero-order integrated rate law. 2.28 M is the concentration using the first-order integrated rate law.

A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 0 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10−4

2.32x

A solution has a [OH−] = 4.2 × 10−8 M at 25 °C. What is the [H3O+] of the solution?

2.4 × 10⁻⁷ M

A reaction follows zero-order kinetics and decays to half of its original concentration of 2.50 M in 53 seconds. What is the value of the rate constant, k?

2.4 × 10−2 M−1∙s−1 [A]t = -kt + [A]0 1.25 M = -k(53 s) + 2.50 M k = 2.4 × 10-2 M·s-1 The rate constant for the reaction is 2.4 × 10−2 M∙s−1. Use the integrated rate law for a zero-order reaction with the given values to determine the rate constant. [A]t = -kt + [A]0 1.25 M = -k(53 s) + 2.50 M k = 2.4 × 10-2 M·s-1 1.3 × 10−2 M∙s−1 is the rate constant if you used the equation for the integrated rate law equation for a first-order reaction. 7.5 × 10−3 M∙s−1 is the rate constant if you used the equation for the integrated rate law equation for a second-order reaction.

Nuclear decay occurs according to first-order kinetics. How long will it take for a sample of radon-218 to decay from 99 grams to 0.50 grams? The half-life of radon-218 is 35 milliseconds.

2.6 × 10^2 ms

What is the pH of a 0.150 M solution of CH3COOH? Ka of CH3COOH = 1.8 × 10−5

2.78

What is the pH of a solution that has 0.125 M CH3COOH and 0.125 M H3BO3? Ka of CH3COOH = 1.8 × 10−5 and Ka of H3BO3 = 5.4 × 10−10

2.82

N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs for 3.5 minutes. If the rate constant, k, equals 5.89 × 10−3, what is the final concentration of N2O5?

2.9 M The final concentration of N2O5 is 2.9 M. Use the integrated rate law for a first-order reaction with the given values to determine the final concentration. ln[A]t = -kt + ln[A]0 ln[A]t = -(5.89×10-3 min-1)(3.5 min) + ln[3.0 M]0 ln[A]t = 1.08 [A]t = 2.9 M 3.0 M is the concentration if the zero-order integrated rate law is used. 2.8 M is the concentration if the second-order integrated rate law is used.

A solution has a [OH−] = 3.4 × 10−5 M at 25 °C. What is the [H3O+] of the solution?

2.9 × 10⁻¹⁰ M

The rate law for the following reaction is rate = k[ClO2]2[OH−]. What is the value of k based on the given data? 2 ClO2(aq) + 2 OH−(aq) → ClO3−(aq) + ClO2−(aq) + H2O(l) Trial [ClO2] (M) [OH-] (M) Initial rate (M/s) 1 0.100 0.100 0.210 2 0.200 0.100 0.840 3 0.200 0.200 1.680

210. M−2 s−1 rate = k[ClO2]2[OH−] 0.210 M/s = k(0.100)2(0.100) k = 210. M−2 s−1 rate = k[ClO2]2[OH−] 0.840 M/s = k(0.200)2(0.100) k = 210. M−2 s−1 rate = k[ClO2]2[OH−] 1.680 M/s = k(0.200)2(0.200) k = 210. M−2 s−1

Identify the balanced nuclear equation for the alpha decay of 220_86Rn

220/86Rn-> 4/2He+216/84Po

Identify the balanced nuclear equation for the alpha decay of 226/88 Ra

226/88Ra -> 4/2He+222/86Rn

The isotope 234/90Th decays by releasing a beta particle. What is the resulting isotope?234/90Th-> 0/-1e+ ?

234/91Pa

Barium-131 is used in the detection of bone tumors. The half-life of barium-131 is approximately 12 days. How long will it take for the radiation level of potassium-131 to drop to 1/4 of its original level?

24 days

Nuclear decay occurs according to first-order kinetics. Scandium-43 decays with a rate constant of 0.178 days−1. After 15.0 days, a sample has a mass of 1.73 g. What was the original mass of the sample?

25.0 g

Based on the data given and a rate constant of 0.031 M−1⋅min−1, calculate the time at which the concentration of reactant A will be 0.100 M . Express your answer in minutes to two significant figures.

260 min Using the integrated rate law for a second-order reaction, the time at which the concentration of reactant A will be 0.100 M can be calculated as t ===(1[A]t−1[A]0)/k(10.100M−10.500 M)/0.031 M−1⋅min−1 258 min

A certain first-order reaction has a rate constant of 7.90×10−3 s−1. How long will it take for the reactant concentration to drop to 18 of its initial value? Express your answer with the appropriate units.

263 s

Given the rate law, rate = k[A][B]3, the rate will increase by a factor of ____ when the concentration of B is tripled.

27 Given the rate law, rate = k[A][B]3, the rate will increase by a factor of 27 when the concentration of B is tripled. If the initial [B] is x, then the final [B] is 3x. Therefore, the rate will increase by a factor of 33, which is 27.

The rate constant for the decomposition of N2O5 is 7.78 × 10−7 at 273 K and 3.46 × 10−5 at T2. If the activation energy is 1027 kJ/mol, what is the final temperature? 2 N2O5(g) → 4 NO2(g) + O2(g)

275 K The final temperature is 275 K. natural log of k2 over k1 equals activation energy over R times the result of 1 over T1 minus 1 over T2 natural log of 3.46 × 10−5 over 7.78 × 10−7 equals 1.027 × 106 Joules per mole divided by 8.314 Joules per mole Kelvin times the result of 1 over 273 Kelvin minus 1 over T2 3.07 × 10−5 equals 1 over 273 Kelvin minus 1 over T2, so T2 equals 275 Kelvin 37.0 K is the temperature if the activation energy is left in kilojoules. 272 K is the temperature if the log is used instead of natural log. 313 K is the temperature if the activation energy and R are inverted.

What is ∆G when ∆G° is 2827 kJ and the pressure of each gas is 0.0391 atm at 25°C? 6 CO2(g) + 6 H2O(g) → C6H12O6(s) + 6 O2(g)

2875 kJ

If two or more chemical reactions can be combined to find an overall reactions, then their equilibrium constants can also be combined to find the equilibrium constant of the overall reaction. Given the following equilibrium reactions and constants, A ⇌ 2 B K1 = 2.4 × 102 B ⇌ C K2 = 0.35 what is the equilibrium constant for the reaction A ⇌ 2 C?

There are six different reactions you can access in the simulation using the drop-down menu. Which of the following are second-order reactions? Check all that apply. 2NO2→2NO+O2 2N2O5→4NO2+O2 2N2O→2N2+O2 C2H6→2CH3 2HI→H2+I2

2NO2→2NO+O2 2HI→H2+I2 Characteristics of second-order reactions For a second-order reaction, [A]→products, the rate of the reaction is given as rate= k[A]2, where k is the rate constant and [A] is the concentration of reactant A. The integrated rate law for second-order reactions is 1[A]t=kt+1[A]0, where [A]t is the concentration of reactant A at time t, k is the rate constant, and [A]0 is the initial concentration of reactant A. This equation is of the type y=mx+b. Therefore, the plot of 1[A]t versus time is always a straight line with a slope k and a y intercept 1[A]0.

Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 2X→Y+Z Y+2L→M+Z What is the overall reaction? Express your answer as a chemical equation. Which species is a reaction intermediate? What is the rate law for step 1 of this reaction?

2X+2L→2Z+M Y Rate = k*[X]^2

A reaction has the general rate law, rate = k[A]2[B]2. What is the order with respect to A and the overall order of the reaction?

2nd order with respect to A; 4th order overall For a reaction with the general rate law, rate = k[A]2[B]2, the reaction is 2nd order with respect to A and 4th order overall. The order with respect to a particular substance is equal to its superscript in the rate law. The overall order for a reaction is the sum of the orders with respect to each reactant.

Nuclear decay occurs according to first-order kinetics. What is the half-life of polonium-218 if a sample decays from 55.4 g to 31.7 g in 2.50 minutes?

3.09 minutes

What is the equilibrium constant, K, for the following reaction at 25°C? MgCO3(s) ⇌ MgO(s) + CO2(g) ΔG° = 48.2 kJ

3.56 × 10−9

A solution has a [Ag+] of 0.00175 M. What concentration of chromate is needed before precipitation begins? Ksp of Ag2CrO4 = 1.12 × 10-12

3.66 × 10-7 M

A reaction mixture initially has [CH3OH] = 0.500 M. What is the equilibrium concentration of CO? CH3OH(g) ⇌ CO(g) + 2 H2(g) K = 4.42 × 10-7 at 125 K

3.81 x 10-3 M K= [CO][H2]^2 / [CH3OH] 4.42 × 10-7 = (X)(2X)^2/ (0.500) 2.21x10^-7 = 4X^3 X= 3.81x10^-3 M

What is the pH of a 0.200 M H2S solution? Ka1 of H2S = 8.9 × 10−8 and Ka2 = 1 × 10−19

3.89

Given the rate law, rate = k[A]3[B]2, the rate will increase by a factor of ____ when the concentration of A and B are both doubled.

32 If the initial [A] is x, then the final [A] is 2x. Therefore, the rate will increase by a factor of 23, which is 8. If the initial [B] is y, then the final [B] is 2y. Therefore, the rate will increase by a factor of 22, which is 4. The total change in rate will increase by a factor of 8 × 4 = 32.

If ∆H = −251 kJ and ∆S = −391 J/K, the spontaneity of the reaction depends on temperature. Below what temperature in Celsius will the reaction be spontaneous?

369 °C

Identify the balanced nuclear equation for the positron emission of 38/19 K.

38/19 K-> 0/1e+38/18 Ar

Given the rate law, rate = k[A]2[B], the rate will increase by a factor of ____ when the concentration of A is doubled.

4 Given the rate law, rate = k[A]2[B], the rate will increase by a factor of 4 when the concentration of A is doubled. If the initial [A] is x, then the final [A] is 2x. Therefore, the rate will increase by a factor of 22, which is 4.

A solution has a [H3O+] = 6.2 × 10−5 M. What is the pH of the solution?

4.21

Nuclear decay occurs according to first order kinetics. How long will it take for a sample of titanium-45 to decay from 65.0 grams to 25.0 grams? The half-life of titanium-45 is 3.08 hours.

4.24 hours

Nuclear decay occurs according to first-order kinetics. What is the half-life for uranium-238, which has a rate constant of 1.54 × 10-10 years-1?

4.50 × 10^9 years

Radioactive decay follows first-order kinetics. If sodium-24 has a half-life of 14.96 hours, what is its rate constant, k?

4.63 × 10^−2 hr−1 The rate constant is 4.63 × 10−2 hr−1. Use the half-life equation for a first-order reaction to find the rate constant from the given half-life. half-life equals 0.693 over k. 14.96 hours equals 0.693 over k, so k equals 4.63 times 10^−2 inverse hours 10.4 hr−1 is the result of multiplying 0.693 times k. 21.6 hr−1 is the result of dividing 14.96 by 0.693.

What is the pH of a 0.110 M solution of HBrO? Ka of HBrO = 2.8 × 10−9

4.74

The reaction of NO and O3 reacts with second order kinetics. If it takes 94 seconds for the concentration of NO to go from 3.00 M to 1.25 M, what is the rate constant, k? NO(g) + O3(g) → NO2(g) + O2(g)

4.96 × 10-3 M−1∙s−1 The rate constant is 4.96 × 10-3 M−1∙s−1. Use the integrated rate law for second-order reactions with the given values to find the rate constant. one over the concentration of A at time t equals kt plus one over the initial concentration of A. one over 1.25 equals k times 94 seconds plus 1 over 3.00 molar. k equals 4.96 × 10−3 inverse molar times inverse seconds 1.86 × 10−2 M−1∙s−1 is the rate constant using the zero-order integrated rate law. 9.31 × 10−3 M−1∙s−1 is the rate constant using the first-order integrated rate law.

Give the complete symbol, with superscript and subscript, for an alpha particle.

4/2He

Give the symbol to complete the balanced nuclear equation.252/99 Es-> ? +248/97Bk

4/2He

If 8.0 mg of a radioactive substance naturally decays to 0.50 mg over 184 days, what is the half-life of the radioisotope?

46 days

The half-life of potassium-42 is approximately 12 hours. How long will it take for the radiation level of potassium-42 to drop to 1/16 of its original level?

48 hours

A 0.200 M solution of a weak acid has a pH of 2.50. What is the value of Ka for the acid?

5.1 × 10-5

A 0.165 M solution of a weak acid has a pH of 3.02. What is the value of Ka for the acid?

5.5 × 10-6

If two or more chemical reactions can be combined to find an overall reaction, then their equilibrium constants can also be combined to find the equilibrium constant of the overall reaction. Given the following equilibrium reactions and constants, A + B ⇌ C K1 = 1.3 × 103 2C ⇌ B + D K2 = 4.3 × 105 what is the equilibrium constant for the reaction A + C ⇌ D?

5.6 × 10^8 K1 * K2 = (1.3 × 103)(4.3 × 105) = 5.6 × 10^8

A solution has a [OH−] = 1.8 × 10−10 M at 25 °C. What is the [H3O+] of the solution?

5.6 × 10⁻⁵ M

What is the pH of a 0.200 M CH3NH3Br solution? Kb of CH3NH2 = 4.4 × 10−4

5.68

The rate law for the following reaction is rate = k[NO][Cl2]. What is the value of k based on the given data? 2 NO(g) + Cl2(g) → 2 NOCl(g) Trial [NO] (M) [Cl2] (M) Initial rate (M/s) 1 0.025 0.030 0.432 2 0.100 0.030 1.728 3 0.100 0.015 0.864

576 M−1 s−1 rate = k[NO][Cl2] 0.432 M/s = k(0.025)(0.030) k = 576 M−1 s−1 rate = k[NO][Cl2] 1.728 M/s = k(0.100)(0.030) k = 576 M−1 s−1 rate = k[NO][Cl2] 0.216 M/s = k(0.100)(0.015) k = 576 M−1 s−1

N2O5 decomposes to form NO2 and O2 with first-order kinetics. How long does it take for the N2O5 concentration to decrease from its initial value of 2.75 M to its final value of 1.85 M, if the rate constant, k, equals 5.89 × 10−3?

67.3 minutes ln[A]t = -kt + ln[A]0 ln[1.85]t = -(5.89×10-3 min-1)t + ln[2.75 M]0 t = 67.3 min The time to decrease from 2.75 M to 1.85 M of N2O5 is 67.3 minutes. Use the integrated rate law for a first-order reaction with the given values to determine the time to reach the final concentration. ln[A]t = -kt + ln[A]0 ln[1.85]t = -(5.89×10-3 min-1)t + ln[2.75 M]0 t = 67.3 min 153 min is the time if the zero-order integrated rate law is used. 30.0 min is the time if the second-order integrated rate law is used.

The rate law for the following reaction is rate = k[NO][NO2]. What is the value of k based on the given data? NO(g) + NO2(g) + O2(g) → N2O5(g) Trial [NO] (M) [NO2] (M) [O2] (M) Initial rate (M/s) 1 0.100 0.050 0.100 0.035 2 0.100 0.100 0.100 0.070 3 0.200 0.050 0.100 0.070 4 0.100 0.050 0.200 0.035

7.0 M−1 s−1 rate = k[NO][NO2] 0.035 M/s = k(0.100)(0.050) k = 7.0 M−1 s−1 rate = k[NO][NO2] 0.070 M/s = k(0.100)(0.100) k = 7.0 M−1 s−1 rate = k[NO][NO2] 0.070 M/s = k(0.200)(0.050) k = 7.0 M−1 s−1 rate = k[NO][NO2] 0.035 M/s = k(0.100)(0.050) k = 7.0 M−1 s−1

A reaction mixture initially has [H2S] = 0.100 M. What is the equilibrium concentration of S2? 2 H2S(g) ⇌ 2 H2(g) + S2(g) K = 1.67 × 10-7 at 298 K

7.47 × 10-4 M K= [H2]^2[S2] / [H2S]^2 1.67x10-7 = (2X)^2(X) / (0.100)^2 1.67x10^-7 = 4X^3 X= 7.47x10^-4 M

What is the pOH for a solution with [OH−] = 3.2 × 10-8 M?

7.49

If two or more chemical reactions can be combined to find an overall reactions, then their equilibrium constants can also be combined to find the equilibrium constant of the overall reaction. Given the following equilibrium reactions and constants, A + B ⇌ C K1 = 0.392 B + D ⇌ C K2 = 0.0483 what is the equilibrium constant for the reaction A ⇌ D?

8.12 K1 * 1/K2 = (0.392) (1/0.0483) = 8.12

A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 21.4 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10−4

8.30 The pH after 21.4 mL of NaOH has been added is 8.30. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCOOH + NaOH → HCOONa + H2O I 3.75 × 10−3 mol 3.75 × 10−3 mol 0 ~ C −3.75 × 10−3 mol −3.75 × 10−3 mol +3.75 × 10−3 mol ~ F 0 0 3.75 × 10−3 mol ~ After the HCOOH and NaOH react, 3.75 × 10−3 moles of HCOO− remain in a total volume of 51.4 mL (30.00 mL HCOOH + 21.4 mL NaOH). Since only HCOO− is in solution, look at the hydrolysis of the cation to determine the pH. Use the moles of HCOO− and the total volume to find the concentration of HCOO− in solution. HCOO− + H2O → HCOOH + OH− I 7.30 × 10−2 M ~ 0 0 C −x ~ +x +x E 7.30 × 10−2 M − x ~ x x The Ka of HCOOH is given but the Kb of HCOO− is needed. Use Ka and Kw to find Kb. Ka × Kb = Kw 1.8 × 10−4 × Kb = 1.0 × 10−14 Kb = 5.6 × 10−11 The Kb can now be used to solve the equilibrium calculation to find [OH−]. Kb equals concentration of HCOOH times concentration of OH negative divided by the concentration of HCOO minus. 5.6 × 10^-11 equals x squared over 7.30 × 10^-2 minus x Assume the x is much less than 7.30 × 10−2. 5.6 × 10^-11 equals x squared over 7.30 × 10^-2, so x equals 2.0 × 10^-6. [OH−] is equal to x so it can be used to find pOH. pOH = −log(2.0 × 10−6) pOH = 5.70 pH + pOH = 14.00 pH + 5.70 = 14.00 pH = 8.30 5.70 is incorrect because it is the pOH, not the pH. 2.44 is incorrect because the Kb was used instead of the Ka. 7.66 is incorrect because it uses the moles of HCOO− instead of the molarity of HCOO−.

What is the pH of a 0.225 M KNO2 solution? Ka of HNO2 = 4.6 × 10−4

8.34 Ka × Kb = Kw (4.6 × 10−4)(Kb) = 1.0 × 10-14 Kb = 2.2 × 10−11 solve for X X= 2.2x10^-6 pOH= -log(2.2x10^-6) pOH= 5.66 pH + pOH = 14 14-5.66= 8.34 pH= 8.34

A substance, Z, with an initial concentration of 1.75 M follows second-order kinetics and decays with a half-life of 6.8 minutes. What is the value of the rate constant, k?

8.4 × 10−2 M−1∙min−1 6.8 min = 1 / k[1.75M]0 K = 8.4 × 10−2 M−1∙min−1 The rate constant is 8.4 × 10−2 M−1∙min−1. Use the half-life equation and the given half-life to solve for the rate constant. 0.13 M−1∙min−1 is the rate constant using the zero-order half-life equation. 0.10 M−1∙min−1 is the rate constant using the first-order half-life equation.

The rate constant for the decomposition of HI is 3.45 × 10−3 at 325 K. What is the rate constant, k2, at 425 K given an activation energy of 592 kJ/mol?

8.44 × 10^19 lnk2 - ln(3.45e-3) = (5.92e5/ 8.314)(1/325 - 1/425) lnK2= 45.88 k2 = 8.44 x 10^19 The rate constant at 425 K is 8.44 × 1019. natural log of k2 over k1 equals activation energy over R times the result of 1 over T1 minus 1 over T2 natural log of k2 minus natural log of k1 equals activation energy over R times the result of 1 over T1 minus 1 over T2 natural log of k2 minus natural log of 3.45 times 10^−3 equals 5.92 times 10^5 Joules per mole divided by 8.314 Joules per mole Kelvin times the result of 1 over 325 Kelvin minus 1 over 425 Kelvin 1.23 × 1049 is the rate constant if the log is used instead of the natural log. 3.63 × 10−3 is the rate constant if the activation energy is left in units of kilojoules.

Radioactive decay follows first-order kinetics. If a sample of I-131 decays from 25.0 mg to 8.25 mg in 12.8 days, what is the rate constant, k?

8.66 × 10−2 day−1 The rate constant is 8.66 × 10−2 day−1. Use the integrated rate law for a first-order reaction with the given values to determine the rate constant. The masses can be used in the equation instead of concentrations. ln[A]t = -kt + ln[A]0 ln[8.25 mg]t = -k(12.8 days) + ln[25.0 mg]0 k = 8.66 × 10-2 day-1 1.31 day−1 is the rate constant if the zero-order integrated rate law is used. 6.34 × 10−3 day−1 is the rate constant if the second-order integrated rate law is used.

A reaction mixture initially has [NO2] = 0.100 M and [ClNO] = 0.100 M. What is the equilibrium concentration of NO? NO2(g) + ClNO(g) ⇌ ClNO2(g) + NO(g) K = 7.7 × 10-5 at 298 K

8.8 × 10-4 M

A substance, A, decays according to zero-order kinetics with a rate constant, k = 4.94 × 10−2 M∙s−1. How long does it take for a 0.750 M sample decay to 0.315 M?

8.81 seconds [A]t = -kt + [A]0 0.315 M = -(4.94×10-2 M·s-1)(t) + 0.750 M t = 8.81 s The time to decrease from 0.750 M to 0.315 M is 8.81 seconds. Use the integrated rate law for a zero-order reaction with the given values to determine the time. [A]t = -kt + [A]0 0.315 M = -(4.94×10-2 M·s-1)(t) + 0.750 M t = 8.81 s 17.6 seconds is the time if you used the equation for the integrated rate law equation for a first-order reaction. 37.3 seconds is the time if you used the equation for the integrated rate law equation for a second-order reaction.

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 10.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

8.96

What is the pOH for a solution with [H3O+] = 1.5 × 10-5 M?

9.18

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 7.50 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

9.26 The pH after 7.50 mL of HCl has been added is 9.26. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCl + NH3 → NH4Cl + H2O I 1.50 × 10−3 mol 3.00 × 10−3 mol 0 ~ C −1.50 × 10−3 mol −1.50 × 10−3 mol +1.50 × 10−3 mol ~ F 0 1.50 × 10−3 mol 1.50 × 10−3 mol ~ After the NH3 and HCl react, 1.50 × 10−3 moles of NH3 and 1.50 × 10−3 moles of NH4+ remain in a total volume of 27.50 mL (7.50 mL HCl + 20.00 mL NH3). Since NH3 and NH4+ are both in solution, the mixture is a buffer, so the Henderson-Hasselbalch equation can be used to solve for the pOH. pOH equals pKb plus log of concentration of BH+ divided by the concentration of B, pOH equals negative log of 1.8 × 10^-5 plus log of 1.50 x 10^-3 divided by 1.50 × 10^-3, pOH equals 4.74 minus 0, so pOH equals 4.74. pH + pOH = 14.00 pH + 4.74 = 14.00 pH = 9.26 4.74 is incorrect because it is the pOH, not the pH. 2.82 is incorrect because it is the pOH when the moles of the acid is used instead of the acid/mole ratio. 11.18 is incorrect because it is the pH when the moles of the acid is used instead of the acid/mole ratio.

Which of the following are correct for zero-order reactions? Check all that apply. The rate of reaction does not equal the rate constant. A higher concentration of reactants will not increase the reaction rate. The units for the rate constant and the rate of reaction are the same. A zero-order reaction slows down as the reaction proceeds. The concentration of the reactants changes nonlinearly.

A higher concentration of reactants will not increase the reaction rate. The units for the rate constant and the rate of reaction are the same. A zero-order reaction is independent of the concentration of the reactants. Zero-order reactions have the following features: The concentration versus time graph is a straight line with a negative slope; this negative slope represents the rate constant of a reaction. The rate of the reaction is equal to the rate constant. The units of k and the rate of reaction are mol⋅L−1⋅s−1 . Photochemical reactions and surface reactions are examples of zero-order reactions.

Which one of the following is a spontaneous process at 25°C?

A spontaneous process at 25°C is ice melting because it occurs without continuous outside intervention. A ball rolling up hill would require outside intervention, such as someone pushing it, so it is not spontaneous. NaCl will dissolve in water at 25°C but will not melt, since its melting point is 801°C, so it is not spontaneous. A canoe going against the flow of the river will require continuous outside intervention, so it is not spontaneous.

A buffer solution has 0.750 M H2CO3 and 0.650 M HCO3−. If 0.020 moles of HCl is added to 275 mL of the buffer solution, what is the pH after the addition? The pKa of carbonic acid is 6.37.

After 0.020 moles of HCl is added to 275 mL of a buffer solution that is 0.750 M H2CO3 and 0.650 M HCO3−, the pH is 6.22. A weak acid dissociates according to the following equation HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq) HCl is a strong acid which will dissociate into H+ and Cl−. The Cl− has no effect on the reaction but the H+ will. The addition of H+ will result in the equilibrium being shifted to the left. Therefore, the moles of HA will increase by the moles of added HCl and the moles of A− will decrease by the same amount. Use the Henderson-Hasselbalch equation to see how the addition of HCl will change the pH. pH equals pKa plus log of the result of the concentration of A minus divided by the concentration of HA pH equals pKa plus log of the result of the concentration of HCO3 minus over the concentration of H2CO3 Since additional acid is being added to the buffer, the amounts of H2CO3 and HCO3− must be substituted into the equation in units of moles. Then the moles of added acid can be added to the H2CO3 and subtracted from HCO3−. 275 milliliters buffer times 1 liter over 1000 milliliters times 0.750 moles H2CO3 over 1 liter solution equals 0.206 moles H2CO3 275 milliliters buffer times 1 liter over 1000 milliliters times 0.650 moles HCO3 minus over 1 liter solution equals 0.179 moles HCO3 minus Substitute the amounts in moles into the Henderson-Hasselbalch equation. pH equals 6.37 plus log of the result of 0.179 moles over 0.206 moles Since both amounts are in the same volume, dividing moles and dividing molarity will result in the same value. Since HCl is being added, the amount of H2CO3 will increase and the amount of HCO3− will decrease so we need to add 0.020 moles to the moles of H2CO3 and subtract 0.020 moles from HCO3−. pH equals 6.37 plus log of open parentheses of the result of 0.179 moles minus 0.020 moles divided by the result of 0.206 moles plus 0.020 moles close parentheses. pH equals 6.37 minus 0.153, which equals 6.22 The answer 6.31 is incorrect because that is the pH of the buffer before the HCl was added. The answer 6.52 is incorrect because the ratio of concentrations is inverted. The answer 6.40 is incorrect because the 0.020 moles was added to the concentration of HCO3− and subtracted from the concentration of H2CO3.

In the reaction shown below, which substance is reduced? Mg0(s) + Ag2+(aq) → Mg2+(aq) + Ag0(s)

Ag2+(aq) is reduced to Ag0(s), as the oxidation state decreases from +2 to 0. Mg0(s) + Ag2+(aq) → Mg2+(aq) + Ag0(s) A decrease in the oxidation state occurs due to reduction, as electrons are gained, while an increase in oxidation state is due to oxidation, as electrons are lost. Mg0(s) is oxidized to Mg2+(aq) in this reaction.

In the titration of a weak base with a strong acid with a 1:1 ratio, when does the pOH equal to pKb?

At half the volume it takes to reach the equivalence point

In the titration of a weak base with a strong acid with a 1:1 ratio, when does the pOH equal to pKb?

At half the volume it takes to reach the equivalence point In the titration of a weak base with a strong acid with a 1:1 ratio, the pOH equals the pKb at half the volume it takes to reach the equivalence point. After adding half of the volume of acid it takes to reach the equivalence point, the moles of acid and base are equal to one another. This simplifies the Henderson-Hasselbalch equation to pOH = pKb because the log of 1 = 0.

Which substance is the oxidizing agent in the following reaction? 2 Au3+(aq) +6 I−(aq) → 2 Au0(s) + 3 I20(s)

Au3+(aq)

Which substance is the oxidizing agent in the following reaction? 2 Au3+(aq) + 6 I−(aq) → 2 Au(s) + 3 I2(s)

Au3+(aq) is the oxidizing agent as it causes I-(aq) to be oxidized to I2(s). I−(aq) is oxidized as the oxidation state of iodine increases from −1 to 0. 2 Au3+(aq) + 6 I−(aq) → 2 Au(s) + 3 I2(s) In the process, Au3+(aq) is reduced to Au(s). The species that is oxidized is the reducing agent, and the species that is reduced is the oxidizing agent.

Predict the type of particle emitted by potassium-40 and why.

Beta; N/Z too high

Predict the type of particle emitted by sodium-24 and why.

Beta; N/Z too high

Predict the type of particle emitted by strontium-90 and why.

Beta; N/Z too high

Which substance is the reducing agent in the following reaction? Co0(s) + Pt2+(aq) → Pt0(s) + Co2+(aq)

Co0(s)

In the reaction shown below, which substance is oxidized? 2 Cr(s) + Fe2O3(aq) → Cr2O3(aq) + 2 Fe(s)

Cr(s)

Select the half-reaction that is properly balanced with respect to oxygen atoms. The half-reaction will not be completely balanced. Cr2O72−(aq) → 2 Cr3+(aq)

Cr2O72−(aq) → 2 Cr3+(aq) + 7 H2O(l)

Which step is the fastest? A⟶B B⟶C C⟶D

C⟶D

Analyzing a specific reaction Consider the following reaction: NO+O3→NO2+O2, rate=k[NO][O3] What would happen to the rate if [NO] were doubled?

Double

Which substance is the reducing agent in the following reaction? Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)

Fe(s) is the reducing agent as it causes CuSO4(aq) to be reduced to Cu0(s). Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq) The oxidation state on the copper atom in CuSO4(aq), is +2, as the charge on the sulfate ion, SO4−2, is −2. The oxidation state on solid copper, Cu(s), is 0. Notice that the copper in CuSO4(aq) is reduced as its oxidation state decreases from +2 to 0. In the process, the Fe(s) atom is oxidized to FeSO4(aq). The oxidation state on solid iron, Fe(s), is 0. The oxidation state on the iron atom in FeSO4 (aq), is +2, as the charge on the sulfate ion, SO4−2, is −2. Therefore, iron is oxidized as the oxidation state increases from 0 to +2. The species that is oxidized is the reducing agent, and the species that is reduced is the oxidizing agent.

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Is the reaction first order or second order with respect to [C12H22O11]?

First order: Either plot a graph or use the equation trial and error method: K1 = 2.303/t * log (a/a-x) Upon graphing a plot of the ln [C12H22O11]ln [C12H22O11] versus time, the fit is linear, so the reaction is first order. A second-order reaction would generate a linear plot of the reciprocal of the concentration versus time.

For the titration of a strong acid with a strong base, the pH at the equivalence point is _________.

For the titration of a strong acid with a strong base, the pH at the equivalence point is neutral. When a strong base reacts with a strong acid, the resulting salt contains the cation of the strong base and the anion of the strong acid. Neither the cation nor the anion will undergo hydrolysis, so the salt solution will be neutral.

What substance is added to balance the number of hydrogen atoms in a half-reaction?

H+ is added to balance the hydrogen in a half-reaction. H2, H2O2, and H2O are incorrect because they are not used to balance the hydrogen in a half-reaction.

Identify the formula for the conjugate acid of HCO3−.

H2CO3

What substance is added to balance the number of oxygen atoms in a half-reaction?

H2O

What is the conjugate base of H3PO4?

H2PO4

Select the strong acid. ANSWER Unselected H2SO4 Unselected H2S Unselected H2SO2 Unselected H2SO3 Unselected I DON'T KNOW YET

H2SO4

Identify the formula for the conjugate acid of H2O.

H3O+

Identify the formula for the conjugate acid of H2PO4−.

H3PO4

When hydrobromic acid (HBr) reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

HBr(aq) + H₂O(l) → H₃O⁺(aq) + Br⁻(aq)

When hydrofluoric acid reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

HF(aq) + H₂O(l) ↔ H₃O⁺(aq) + F⁻(aq)

Identify the formula for the conjugate acid of NO3−.

HNO3

Select the strong acid. HNO2 HCN NH3 HNO3 I DON'T KNOW YET

HNO3

When nitrous acid reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

HNO₂(aq) + H₂O(l) → H₃O⁺(aq) + NO₂⁻(aq)

Identify the formula for the conjugate base of H2SO3.

HSO3−

Identify the formula for the conjugate acid of SO42−.

HSO4−

When a hydrogen sulfate ion reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

HSO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + SO₄²⁻(aq)

When a hydrogen sulfide ion reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

HS⁻(aq) + H₂O(l) → H₃O⁺(aq) + S²⁻(aq)

Identify the formula for the conjugate base of H2S.

HS−

Which best defines half-life?

Half-life is the time it takes for the amount of a substance to decrease to half of its original value. Half-life is the time it takes for the amount of a substance to decrease to half of its original value. In one half-life, a sample decreases to half of its original concentration. After a second half-life, the sample has decreased to one-fourth of its original concentration. The amount of the sample continues to decrease by half for each successive half-life.

Which substance is the oxidizing agent in the following reaction? Mg(s) +2 HCl(aq) → H2(g) + MgCl2(aq)

Hydrogen in HCl(aq) is the oxidizing agent as it causes Mg(s) to be oxidized to MgCl2(aq) The oxidation state on solid magnesium, Mg(s), is 0, while the oxidation state on magnesium in MgCl2(aq) is +2. Notice that Mg(s) is oxidized as the oxidation state increases from 0 to +2. In the process, the hydrogen in HCl(aq) is reduced to H2(g). The oxidation state on hydrogen gas, H2(g), is 0, while the oxidation state on the hydrogen atom in HCl(aq) is +1. Therefore, hydrogen is reduced as the oxidation state decreases from +1 to 0. The species that is oxidized is the reducing agent, and the species that is reduced is the oxidizing agent.

When dihydrogen phosphate ion reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentration of A− would decrease. If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentration of HA would decrease. If you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase. The equilibrium constant for this reaction changes as the pH changes.

If you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase. If you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase. TRUE, equilibrium will shift to left, therefore H+ is decreased, therefore pH increases If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentration of HA would decrease. FALSE, it will increase due to the shift of equilibrium If you add the soluble salt KA to a solution of HA that is at equilibrium, the concentration of A− would decrease. FALSE, since you are adding KA--> K+ and A- The equilibrium constant for this reaction changes as the pH changes. Ka wont change with ´pH

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Using the rate constant found in Part B, calculate the concentration of sucrose at 39 min if the initial sucrose concentration were 0.316 M and the reaction were zero order in sucrose. Express your answers using two significant figures.

In zero-order reactions, the rate does not change and the plot of concentration versus time will be linear. Therefore, all of the sucrose will react until it is consumed. The integrated rate law for a zero-order reaction is as follows, where [A]t is the concentration at a given time, k is the rate constant, t is a given time, and [A]0 is the initial concentration. [A]t=−kt+[A]0 The initial concentration of sucrose is 0.316 M , and the rate constant found in Part B is 3.67×10−3 min−1 . Substituting these values into the integrated rate law gives the concentration of sucrose remaining after 39 min : [sucrose]39 min=−(3.67×10−3 min−1)(39 min)+0.316 M[sucrose]39 min=0.17 M

What is the equilibrium constant, K, for the following reaction, if the concentration of CO was measured to be 3.1 × 10-2 M, the concentration of O2 was measured to be 1.7 × 10-2 M, and CO2 was measured to be 1.452 × 10-1 M? 2 CO(g) + O2(g) ⇌ 2 CO2(g)

K = 1.3 × 10^3 K= [1.452e-1]2 / [3.1e-2]2[1.7e-2] K= 1.3e3

What is the equilibrium constant, K, for the following reaction, if the concentration of N2 was measured to be 0.521 M, H2 was measured to be 0.482 M, and NH3 was measured to be 3.17 M? N2(g) + 3 H2(g) ⇌ 2 NH3(g)

K = 172 K= [3.17]2 / [0.521][0.482]3 K= 172

What is the equilibrium constant, K, for the following reaction, if the concentration of N2O4 was measured to be 2.35 × 10-2 M, and NO2 was measured to be 1.3 × 10-2 M? N2O4(g) ⇌ 2 NO2(g)

K = 7.2 × 10-3 K= [1.3e-2]2 / [2.35e-2] K= 7.2 x 10^-3

What is the equilibrium constant, K, for the following reaction, if the concentration of PCl3 and Cl2 was measured to be 6.4 M, and PCl5 was measured to be 3.93 × 10-2 M? PCl3(g) + Cl2(g) ⇌ PCl5(g)

K = 9.6 × 10-4

What is the equilibrium constant expression for the given reaction? PCl3(l) + Cl2(g) ⇌ PCl5(s)

K= 1 / [Cl2]

What is the equilibrium constant expression for the given reaction? CO2(g) + C(s) ⇌ 2 CO(g)

K= [CO]2 / [CO2]

What is the equilibrium constant expression for the following reaction? 2 H2(g) + O2(g) ⇌ 2 H2O(g)

K= [H2O]2 / [H2]2[O2]

What is the equilibrium constant expression for the given reaction? Fe3O4(s) + 4 H2(g) ⇌ 3 Fe(s) + 4 H2O(g)

K= [H2O]4 / [H2]4

What is the equilibrium constant expression for the following reaction? 2 Fe(s) + 3 H2O(g) ⇌ Fe2O3(s) + 3 H2(g)

K= [H2]3 / [H2O]3

What is the equilibrium constant expression for the following reaction? H2(g) + Cl2(g) ⇌ 2 HCl(g)

K= [HCl]2 / [H2][Cl2]

What is the equilibrium constant expression for the following reaction? 3 O2(g) ⇌ 2 O3(g)

K= [O3]2 / [O2]3

What is the equilibrium constant expression for the following reaction? Si(s) + 2 Cl2(g) ⇌ SiCl4(g)

K= [SiCl4] / [Cl2]2

A 0.250 M solution of a weak acid has a pH of 2.67. What is the value of Ka for the acid?

Ka= 1.8x10^-5 pH = -log[H+] 2.67 = -log[H+] 10^-2.67= 2.1x10^-3 Ka= [H+] [A-]/ [HA] Ka= (2.1x10^-3)(2.1x10^-3)/ (0.250) Ka= 1.8x10^-6

What is the expression for Kp for the given reaction? 2 NO2(g) ⇌ N2O4(g)

Kp= PN2O4 / P^2NO2

What is the expression for Kp for the given reaction? N2(g) + 3 H2(g) ⇌ 2 NH3(g)

Kp= [P^2NH3] / [PN2][P^3H2]

Ionic solids dissolve in water and break up into their ions. However, some ionic solids only partially dissolve, leaving a significant amount of solid undissolved. In some cases, the amount that dissolves is very small and is almost zero. A mathematical formula that indicates the extent to which an ionic solid dissolves in water is called the solubility product constant. The solubility product constant, commonly referred to as Ksp, indicates the extent to which an ionic solid dissolves. Which of the following is the correct solubility product constant for the reaction shown below? CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2

Note 1:

Learning Goal: To calculate average and relative reaction rates. You can measure the rate of a reaction, just like you can measure the speed a jogger runs. While a jogger would be reported to run a specific number of miles in an hour, miles/hourmiles/hour, a reaction is reported to form product or consume reagent in molar concentration per second, M/s. Reaction rate can be defined either as the increase in the concentration of a product per unit time or as the decrease of the concentration of a reactant per unit time. By definition, reaction rate is a positive quantity. In the reaction X→2, for example, Y is being produced twice as fast as X is consumed and thus rate of X=1/2(rate of Y) Each rate can be expressed as the change in concentration over the change in time, ΔtDeltat: −Δ[X]/Δ t= 1/2(Δ[Y]/Δt)

Consider the reaction 2H3PO4→P2O5+3H2O Time (s) 0 10.0 20.0 30.0 40.0 50.0 [P2O5] (M) 0 2.70×10−3 5.70×10−3 7.50×10−3 8.70×10−3 9.30×10−3 Determine the average rate of decomposition of H3PO4 between 10.0 and 40.0 s. Express your answer with the appropriate units.

Looking at the equation, rate H3PO4 = 2*(rate of P2O5). 2 * (0.0002 M/s) = 0.0004 M/s

What are the units for the rate constant, k, for a second-order reaction?

M−1∙s−1 The units for the rate constant, k, for a second-order reaction is M−1∙s−1. M∙s−1 are the units for the rate constant, k, for a zero-order reaction. s−1 is the unit for the rate constant, k, for a first-order reaction.

Analyzing a specific reaction Consider the following reaction: NO+O3→NO2+O2, rate=k[NO][O3] What are the units of the rate constant k for this reaction?

M−1⋅s−1 Overall reaction order and rate-constant units The sum of the individual orders gives the overall reaction order. The example equation above is third order overall because 0+1+2=3. For the units of rate to come out to be M/s, the units of the rate constant for third-order reactions must be M−2⋅s−1 since M/s=(M−2⋅s−1)(M3) For a second-order reaction, the rate constant has units of M−1⋅s−1 because M/s=(M−1⋅s−1)(M2). In a first-order reaction, the rate constant has the units s−1 because M/s=(s−1)(M1).

Identify the formula for the conjugate base of NH4+.

NH3

When an ammonium ion reacts with water, water acts as a Brønsted-Lowry base. Choose the reaction that describes this.

NH₄⁺(aq) + H₂O(l) → H₃O⁺(aq) + NH₃(aq)

Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?

NaH2PO4 and Na2HPO4 The pair that can create a buffer solution is NaH2PO4 and Na2HPO4. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, both compounds will completely dissociate and result in H2PO4− and HPO42− ions, which differ by a single proton, H+. Therefore, these two compounds will form a conjugate acid-base pair. NaOH and H2O cannot form a buffer because NaOH is a strong base. Additionally, the two substances are not conjugates of one another. NaCl and KCl cannot form a buffer because they are both neutral salts and are not a conjugate acid-base pair. NH3 and H2O cannot form a buffer because while NH3 is a weak base, H2O is not its conjugate acid.

For a reaction, ∆H is positive and ∆S is negative. Which choice best describes when the reaction will be spontaneous?

Never

In the reaction shown below, which substance is reduced? Fe0(s) + Ni3+(aq) → Fe3+(aq) + Ni0(s)

Ni3+(aq) is reduced to Ni0(s), as the oxidation state decreases from +3 to 0. Fe0(s) + Ni3+(aq) → Fe3+(aq) + Ni0(s) A decrease in the oxidation state occurs due to reduction, as electrons are gained, while an increase in oxidation state is due to oxidation, as electrons are lost. Fe0(s) is oxidized to Fe3+(aq) in this reaction.

If the volume of the reaction vessel is decreased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? Br2(g) + Cl2(g) ⇌ 2 BrCl(g)

No shift

Equilibrium reactions respond to outside stresses according to Le Châtelier's Principle, which states that an equilibrium system will shift in either the forward or reverse direction in order to reduce the stress. Outside stresses include changes in concentration of either the reactants or products, changes in temperature, as well as changes in volume. Changes in volume inversely affect the pressure of the system. If a catalyst is added to the following equilibrium system, in which direction will the reaction shift to relieve the stress? 3 O2(g) + heat ⇌ 2 O3(g)

No shift Why? When a catalyst is added to the equilibrium system, there is no effect on the equilibrium system.

Which substance is the oxidizing agent in the following reaction? 2 NH3(aq) + OCl−(aq) → N2H4(aq) + H2O(l) + Cl−(aq)

OCl−(aq)

What equation best represents the behavior of PbSO4 in water?

PbSO4(s) ⇌ Pb2+(aq) + SO42−(aq)

Predict the type of particle emitted by boron-8 and why.

Positron; N/Z too low

Predict the type of particle emitted by magnesium-23 and why.

Positron; N/Z too low

Predict the type of particle emitted by manganese-50 and why.

Positron; N/Z too low

Predict the type of particle emitted by silicon-27 and why.

Positron; N/Z too low

What is the rate law for the following mechanism in terms of the overall rate constant k? Step1: A+B ⇌ C :Step2: B+C -> D

Rate = k[A][B]2

Consider the reaction 5Br−(aq)+BrO−3(aq)+6H+(aq)→3Br2(aq)+3H2O(l) The average rate of consumption of Br− is 1.06×10−4 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval? Express your answer with the appropriate units.

Rate of formation of Br2 = 6.36×10−5 Ms

Rate of formation of H = double rate of formation Br2 6.36*10-5 M/s * 2 = Rate of consumption of H+ = 1.27×10−4 Ms

Consider the following equilibrium reaction and equilibrium constant: N2(g) + O2(g) ⇌ 2 NO(g) K = 2 × 10-9 Is this equilibrium reaction a product favored equilibrium with mostly products, a reactant favored equilibrium with mostly reactants, or a true equilibrium with both reactants and products?

Reactant favored equilibrium with mostly reactants

If the volume is decreased, resulting in an increase in pressure in the equilibrium system, in which direction will the reaction shift to relieve the stress? C2H2(g) + 2 H2(g) ⇌ C2H6(g)

Shift to the Products

If the concentration of O2 is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g)

Shift to the products

If the temperature is decreased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? H2(g) + Cl2(g) ⇌ 2 HCl(g) + heat

Shift to the products

If the temperature is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? C(s) + H2O(g) + heat ⇌ CO(g) + H2(g)

Shift to the products

If the volume is increased, resulting in a decrease in pressure in the equilibrium system, in which direction will the reaction shift to relieve the stress? 2 NOCl(g) ⇌ Cl2(g) + 2 NO(g)

Shift to the products

If the volume of the reaction vessel is decreased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? CO2(g) + H2(g) ⇌ CO(g) + H2O(l)

Shift to the products

If the volume of the reaction vessel is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? C(s) + CO2(g) ⇌ 2 CO(g)

Shift to the products

If the volume of the reaction vessel is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? CaCO3(s) ⇌ CaO(s) + CO2(g)

Shift to the products

If the concentration of Br2 is decreased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? Br2(g) + Cl2(g) ⇌ 2 BrCl(g)

Shift to the reactants

If the concentration of H2 is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? 2 CH4(g) ⇌ C2H2(g) + 3 H2(g)

Shift to the reactants

If the temperature is decreased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? N2(g) + O2(g) + heat ⇌ 2 NO(g)

Shift to the reactants

If the temperature is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? C2H2(g) + 2 Cl2(g) ⇌ C2H2Cl4(g) + heat

Shift to the reactants

If the temperature is increased in the following equilibrium system, in which direction will the reaction shift to relieve the stress? H2(g) + Cl2(g) ⇌ 2 HCl(g) + heat

Shift to the reactants

Note 1: Learning Goal: To understand how to use integrated rate laws to solve for concentration. A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours? This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110 miles=milemarker 35 If we were to write a formula for this calculation, we might express it as follows: milemarker=milemarker0−(speed×time) where milemarker is the current milemarker and milemarker0 is the initial milemarker.

Similarly, the integrated rate law for a zero-order reaction is expressed as follows: [A]=[A]0−rate×time or [A]=[A]0−kt since rate=k[A]0=k A zero-order reaction (Figure 1)proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates (Figure 2) do change over time as the reactant concentration changes. Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction. The integrated rate law for a first-order reaction is expressed as follows: [A]=[A]0e−kt where k is the rate constant for this reaction. The integrated rate law for a second-order reaction is expressed as follows: 1[A]=kt+1[A]0 where k is the rate constant for this reaction.

What is the Ksp for the following reaction, if the concentration of Ba2+ was measured to be 0.0075 M and F− was measured to be 0.015 M? BaF2(s) ⇌ Ba2+(aq) + 2 F−(aq)

The Ksp for BaF2(s) is calculated to be 1.7 × 10-6, which is shown below. BaF2(s) ⇌ Ba2+(aq) + 2 F−(aq) Ksp = [0.0075][0.015]2 Ksp = 1.7 × 10-6 previous

What is the Ksp for the following reaction, if the concentrations of Ca2+ and SO42− were both measured to be 0.0048 M? CaSO4(s) ⇌ Ca2+(aq) + SO42−(aq)

The Ksp for CaSO4(s) is calculated to be 2.3 × 10-5, which is shown below. CaSO4(s) ⇌ Ca2+(aq) + SO42−(aq) Ksp = [0.0048][0.0048] Ksp = 2.3 × 10-5

What is the Ksp for the following reaction, if the concentration of Fe2+ was measured to be 5.8 x 10-6 M and OH− was measured to be 1.16 x 10-5 M? Fe(OH)2(s) ⇌ Fe2+(aq) + 2 OH−(aq)

The Ksp for Fe(OH)2(s) is calculated to be 7.8 × 10-16, which is shown below. Fe(OH)2(s) ⇌ Fe2+(aq) + 2 OH−(aq) Ksp = [5.8 × 10-6][1.16 × 10-5]2 Ksp = 7.8 × 10-16

What is the Ksp for the following reaction if the concentration of Pb2+ was measured to be 0.0043 M and F− was measured to be 0.0029 M? PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)

The Ksp for PbF2(s) is calculated to be 3.6 × 10-8, which is shown below. PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq) Ksp = [Pb2+][F−]2 Ksp = [0.0043][0.0029]2 Ksp = 3.6 × 10-8

Assume you are in the process of balancing the half-reaction listed below. At this point, all of the atoms are balanced, but the charge needs to be balanced. You will complete this step by adding electrons to one side of the equation. 10 H+(aq) + NO3−(aq)→ NH4+(aq) + 3 H2O(l) Select the half-reaction that is properly balanced with respect to total charge.

The balanced half reaction is 10 H+(aq)+ NO3−(aq) + 8 e− → NH4+(aq) + 3 H2O(l). In order to balance the half-reaction, the total charge on both sides of the equation must be equal. At this point, the total charge on the left side of the equation is +9, while the total charge on the right side of the equation is +1. 10 H+(aq) + NO3−(aq) → NH4+(aq) + 3 H2O(l) Therefore, eight electrons need to be added to the left side of the equation. This now results in a total charge of +1 on the left side, which equals the total charge on the right side of the half-reaction. 10 H+(aq) + NO3−(aq) + 8 e− → NH4+(aq) + 3 H2O(l) 10 H+(aq) + NO3−(aq) + 9 e− → NH4+(aq)+ 3 H2O(l) has the wrong number of electrons. With nine electrons, the total charge on the left is zero, and the charge on the right is +1. 10 H+(aq) + NO3−(aq) → NH4+(aq)+ 3 H2O(l) + 9 e− has the incorrect number of electrons on the wrong side of the equation. As written, the total charge on the left side of the equation is +9, and the total charge on the right side of the equation is −8. 10 H+(aq) + NO3−(aq) → NH4+(aq) + 3 H2O(l) + 8 e− has the electrons on the wrong side of the equation. As written, the total charge on the left side of the equation is +9, and the total charge on the right side of the equation is −7.

What is the balanced half-reaction in acidic aqueous solution for MnO4−(aq) → Mn2+(aq)?

The balanced half-reaction for MnO4−(aq) → Mn2+(aq) is MnO4−(aq) + 8 H+(aq) + 5 e− → Mn2+(aq) + 4 H2O(l). A balanced half-reaction must have all of the elements balanced as well as a balance of total charge on both sides of the equation. MnO4−(aq) + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) + 5 e− is incorrect because the charges are unbalanced. MnO4−(aq) → Mn2+(aq) + ½ O2(g) + 6 e− is incorrect because the oxygen should be balanced with H2O instead of O2, and the charges are not balanced. MnO4−(aq) → Mn2+(aq) + 6 e- is incorrect because not all elements are balanced.

Identify the balanced nuclear equation for the beta decay of zinc-71

The balanced nuclear equation for the beta decay of is: Beta (β) emission is the conversion of an unstable neutron in the nucleus to a proton and an electron. The electron () is ejected as a beta (β) particle, and the proton remains in the nucleus. Since the proton remains in the nucleus, the number of protons increases by one, while the number of neutrons decreases by one. The mass number of the radioactive nucleus and the mass number of the new nucleus are the same. The atomic number of the atom increases by one since there is a new proton. This is a transmutation, resulting in a nucleus of a different element. For this example, the mass number of zinc, 71, is equal to the sum of the mass numbers of the beta particle and the new nucleus (71+0=71). 71 = 0 + ? 71 - 0 = ? 71 - 0 = 71 (mass number of new nucleus) The atomic number of zinc, 30, is equal to the sum of the atomic numbers of the beta particle and the new nucleus (−1+31=30). 30 = -1 + ? 30 + 1 = ? 30 + 1 = 31 (atomic number of new nucleus) On the periodic table, the element with the atomic number 31 is gallium (Ga). The atomic symbol of the new element is . This answer is incorrect since the symbol, , is the symbol for a positron. This answer is incorrect since the symbol, , is the symbol for an alpha particle. This answer is incorrect since the beta particle is shown as a reactant in this equation. In the beta decay of a radioactive nucleus, the beta particle is a product in the balanced nuclear equation.

What is the balanced oxidation half-reaction for the unbalanced oxidation-reduction reaction? Na(s) + Cl2(g) → NaCl(s)

The balanced oxidation half-reaction is Na(s) → Na+(s) + e− for the unbalanced oxidation-reduction reaction, Na(s) + Cl2(g) → NaCl(s). Cl2(g) → 2 e− + 2 Cl−(s) is incorrect because the equation charges are not balanced and the chlorine is reduced in the overall reaction. Cl2(g) + 2 e− → 2 Cl−(s) is incorrect because the chlorine is reduced in the overall reaction. Na(s) + e− → Na+(s) is incorrect because the charge is not balanced. The electron should be on the right side of the equation.

What is the balanced reduction half-reaction for the unbalanced oxidation-reduction reaction? Na(s) + Cl2(g) → NaCl(s)

The balanced reduction half-reaction is Cl2(g) + 2 e− → 2 Cl−(s) for the unbalanced oxidation-reduction reaction, Na(s) + Cl2(g) → NaCl(s). Cl2(g) → 2 e− + 2 Cl−(s) is incorrect because the charges are not balanced. The electrons should be on the left side of the equation. Na(s) + e− → Na+(s) is incorrect because the charges are not balanced and the sodium is oxidized in the overall reaction. Na(s) → Na+(s) + e− is incorrect because the sodium is oxidized in the overall reaction.

Select the best buffer to maintain a solution at pH = 3.5. HCOOH with HCOONa, Ka of HCOOH = 1.8 x 10−4 Unselected HClO with NaClO, Ka of HClO = 3.5 × 10−8 Unselected H3AsO4 with NaH2AsO4, Ka of H3AsO4 = 5.0 × 10−3

The best buffer to maintain at pH = 3.5 is HCOOH with HCOONa, Ka of HCOOH = 1.8 x 10−4. This is the best buffer solution because the pKa = 3.74, so it is closest to the desired pH value and within the ±1 range for an effective buffer. The other options have a pKa that is greater than or less than 1 away from the desired pH value. For HClO with NaClO, Ka of HClO = 3.5 × 10−8, pKa = 7.46. For H3AsO4 with NaH2AsO4, Ka of H3AsO4 = 5.0 × 10−3, pKa = 2.30.

Select the best buffer to maintain a solution at pH = 9.0. HBrO with NaBrO, Ka of HBrO = 2.0 × 10−9 Unselected NaHsAsO4 with Na2HAsO4, Ka of NaH2AsO4 = 9.3 × 10−8 Unselected HClO with NaClO, Ka of HClO = 3.5 × 10−8

The best buffer to maintain at pH = 9.0 is HBrO with NaBrO, Ka of HBrO = 2.0 × 10−9. This is the best buffer solution because the pKa = 8.70, so it is closest to the desired pH value and within the ±1 range for an effective buffer. The other options have a pKa that is greater than or less than 1 away from the desired pH value. For HClO with NaClO, Ka of HClO = 3.5 × 10−8, pKa = 7.46. For NaHsAsO4 with Na2HAsO4, Ka of NaH2AsO4 = 9.3 × 10−8, pKa = 7.03. previous

What equation best represents the behavior of BaSO4 in water?

The best equation to represent the behavior of solid BaSO4 in water is BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq). BaSO4(aq) ⇌ Ba2+(aq) + SO42−(aq) and BaSO4(aq) → Ba2+(aq) + SO42−(aq) show BaSO4 initially as aqueous, but it is a solid when added to the water. BaSO4(s) → Ba2+(aq) + SO42−(aq) and BaSO4(aq) → Ba2+(aq) + SO42−(aq) use a simple arrow in the reaction when an equilibrium arrow should be used because BaSO4 is only slightly soluble in water.

What equation best represents the behavior of solid Mg(OH)2 when added to water?

The best equation to represent the behavior of solid Mg(OH)2 in water is Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH−(aq). Mg(OH)2(aq) ⇌ Mg2+(aq) + 2 OH−(aq) and Mg(OH)2(aq) → Mg2+(aq) + 2 OH−(aq) show Mg(OH)2 initially as aqueous, but it is a solid when added to the water. Mg(OH)2(s) → Mg2+(aq) + 2 OH−(aq) and Mg(OH)2(aq) → Mg2+(aq) + 2 OH−(aq) use a simple arrow in the reaction. Since Mg(OH)2 is only slightly soluble in water, the reaction is an equilibrium so an equilibrium arrow should be used in the equation.

What is the coefficient for H2O(l) when H2O2(aq) + Cr2O72−(aq) → O2(g) + Cr3+(aq) is balanced in acidic aqueous solution?

The coefficient for H2O(l) when H2O2(aq) + Cr2O72−(aq) → O2(g) + Cr3+(aq) is balanced in acidic aqueous solution is 7. Write the two half-reactions for the equation. H2O2(aq) → O2(g) Cr2O72−(aq) → Cr3+(aq) Balance the two half-reactions ((a) balance Cr, (b) add H2O to balance oxygen,(c) add H+ to balance hydrogen, (d) add electrons to balance charge). H2O2(aq) → O2(g) + 2 H+(aq) +2 e− Cr2O72−(aq) + 14 H+(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l) Multiply the half-reactions so the number of electrons is equal. 3 H2O2(aq) → 3 O2(g) + 6 H+(aq) +6 e− Cr2O72−(aq) + 14 H+(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l) Add the half-reactions together. 3 H2O2(aq) + Cr2O72−(aq) + 8 H+(aq) → 3 O2(g) + 2 Cr3+(aq) + 7 H2O(l)

What is the coefficient for H2O(l) when MnO4−(aq) + H2S(g) → S(s) + MnO(s) is balanced in acidic aqueous solution?

The coefficient for H2O(l) when MnO4−(aq) + H2S(g) → S(s) + MnO(s) is balanced in acidic aqueous solution is 6. Write the two half-reactions for the equation. MnO4−(aq) → MnO(s) H2S(g) → S(s) Balance the two half-reactions ((a) balance Mn and S, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen, (d) add electrons to balance charge). MnO4−(aq) + 6 H+(aq) + 5 e−→ MnO(aq) + 3 H2O(l) H2S(g) → S(s) + 2 H+(aq) + 2 e− Multiply the half-reactions so the number of electrons is equal. 2 MnO4−(aq) + 12 H+(aq) + 10 e− → 2 MnO(s) + 6 H2O(l) 5 H2S(g) → 5 S(s) + 10 H+(aq) + 10 e− Add the half-reactions together. 2 MnO4−(aq) + 2 H+(aq) + 5 H2S(g) → 2 MnO(s) + 6 H2O(l) + 5 S(s)

What is the coefficient for H2O(l) when NH3(g) + NO2(g) → N2(g) is balanced in basic aqueous solution?

The coefficient for H2O(l) when NH3(g) + NO2(g) → N2(g) is balanced in basic aqueous solution is 12. 1. Write the two unbalanced half-reactions for the equation. NH3(g) → N2(g) NO2(g) → N2(g) 2. Balance the two half-reactions ((a) balance N, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen). 2 NH3(g) → N2(g) + 6 H+(aq) 2 NO2(g) + 8 H+(aq) → N2(g) + 4 H2O(l) 3. Adjust the two half-reactions for basic solution by adding the enough OH- to both sides of each equation to remove all H+. 2 NH3(g) + 6 OH-(aq) → N2(g) + 6 H+(aq) + 6 OH-(aq) 2 NO2(g) + 8 H+(aq) + 8 OH-(aq) → N2(g) + 4 H2O(l) + 8 OH-(aq) 4. Allow H+ and OH- to "react" to form H2O if they are on the same side of a half-reaction. 2 NH3(g) + 6 OH-(aq) → N2(g) + 6 H2O(l) 2 NO2(g) + 8 H2O(l) → N2(g) + 4 H2O(l) + 8 OH-(aq) 5. Cancel H2O molecules if they appear on opposite sides of the SAME half-reaction. 2 NH3(g) + 6 OH-(aq) → N2(g) + 6 H2O(l) 2 NO2(g) + 4 H2O(l) → N2(g) + 8 OH-(aq) 6. Add electrons to either side balance charge. 2 NH3(g) + 6 OH-(aq) → N2(g) + 6 H2O(l) + 6 e− 2 NO2(g) + 4 H2O(l) + 8 e− → N2(g) + 8 OH-(aq) 7. Multiply the half-reactions so the number of electrons is equal. 8 NH3(g) + 24 OH-(aq) → 4 N2(g) + 24 H2O(l) + 24 e− 6 NO2(g) + 12 H2O(l) + 24 e− → 3 N2(g) + 24 OH-(aq) 8. Add the half-reactions together and cancel any species that are present on both sides of the reaction. 8 NH3(g) + 6 NO2(g) → 7 N2(g) + 12 H2O(l)

What is the coefficient for H2O(l) when PbO2(s) + I2(s) → Pb2+(aq) + IO3−(aq) is balanced in acidic aqueous solution?

The coefficient for H2O(l) when PbO2(s) + I2(s) → Pb2+(aq) + IO3−(aq) is balanced in acidic aqueous solution is 4. Write the two half-reactions for the equation. PbO2(s) → Pb2+(aq) I2(s) → IO3−(aq) Balance the two half-reactions ((a) balance Pb and I, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen, (d) add electrons to balance charge). PbO2(s) + 4 H+(aq) + 2 e− → Pb2+(aq) + 2 H2O(l) I2(s) + 6 H2O(l) → 2 IO3−(aq) + 12 H+(aq) + 10 e− Multiply the half-reactions so the number of electrons is equal. 5 PbO2(s) + 20 H+(aq) + 10 e− → 5 Pb2+(aq) + 10 H2O(l) I2(s) + 6 H2O(l) → 2 IO3−(aq) + 12 H+(aq) + 10 e− Add the half-reactions together. 5 PbO2(s) + 8 H+(aq) + I2(s) → 5 Pb2+(aq) + 4 H2O(l) + 2 IO3−(aq)

What is the coefficient for H2O(l) when S2−(aq) + NO3−(aq) → S(s) + NO(g) is balanced in acidic aqueous solution?

The coefficient for H2O(l) when S2−(aq) + NO3−(aq) → S(s) + NO(g) is balanced in acidic aqueous solution is 4. Write the two half-reactions for the equation. S2−(aq) → S(s) NO3−(aq) → NO(g) Balance the two half-reactions ((a) balance S and N, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen, (d) add electrons to balance charge). S2−(aq) → S(s) + 2 e− NO3−(aq) + 4 H+(aq) + 3 e− → NO(g) + 2 H2O(l) Multiply the half-reactions so the number of electrons is equal. 3 S2−(aq) → 3 S(s) + 6 e− 2 NO3−(aq) + 8 H+(aq) + 6 e− → 2 NO(g) + 4 H2O(l) Add the half-reactions together. 3 S2−(aq) + 2 NO3−(aq)+ 8 H+(aq) → 3 S(s) + 2 NO(g) + 4 H2O(l)

What is the coefficient for OH−(aq) when S2−(aq) + NO3−(aq) → S(s) + NO(g) is balanced in basic aqueous solution?

The coefficient for OH− when S2−(aq) + NO3−(aq) → S(s) + NO(g) is balanced in basic aqueous solution is 8. Write the two half-reactions for the equation. S2−(aq) → S(s) NO3− (aq)→ NO(g) Balance the two half-reactions ((a) balance S and N, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen, (d) add electrons to balance charge). S2−(aq) → S(s) + 2 e− NO3−(aq) + 4 H+(aq) + 3 e− → NO(g) + 2 H2O(l) Multiply the half-reactions so the number of electrons is equal. 3 S2−(aq) → 3 S(s) + 6 e− 2 NO3−(aq) + 8 H+(aq) + 6 e− → 2 NO(g) + 4 H2O(l) Add the half-reactions together. 3 S2−(aq) + 2 NO3−(aq) + 8 H+(aq) → 3 S(s) + 2 NO(g) + 4 H2O(l) To balance in basic aqueous solution, look at the coefficient in front of H+ and add an equal number of OH− ions to each side of the equation. Then simplify so that H2O only appears on one side of the equation. 3 S2−(aq) + 2 NO3−(aq) + 8 H+(aq) + 8 OH−(aq) → 3 S(s) + 2 NO(g) + 4 H2O(l) + 8 OH−(aq) 3 S2−(aq) + 2 NO3−(aq) + 8 H2O(l) → 3 S(s) + 2 NO(g) + 4 H2O(l) + 8 OH−(aq) 3 S2−(aq) + 2 NO3−(aq) + 4 H2O(l) → 3 S(s) + 2 NO(g) + 8 OH−(aq) previous next

What is the coefficient for OH−(aq) when H2O2(aq) + Cr2O72−(aq)→ O2(g) + Cr3+(aq) is balanced in basic aqueous solution?

The coefficient for OH−(aq) when H2O2(aq) + Cr2O72−(aq)→ O2(g) + Cr3+(aq) is balanced in basic aqueous solution is 8. 1. Write the two unbalanced half-reactions for the equation. H2O2(aq) → O2(g) Cr2O72−(aq) → Cr3+(aq) 2. Balance the two half-reactions ((a) balance Cr (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen). H2O2(aq) → O2(g) + 2 H+(aq) Cr2O72−(aq) + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(l) 3. Adjust the two half-reactions for basic solution by adding the enough OH- to both sides of each equation to remove all H+. H2O2(aq) + 2 OH-(aq) → O2(g) + 2 H+(aq) + 2 OH-(aq) Cr2O72−(aq) + 14 H+(aq) + 14 OH-(aq) → 2 Cr3+(aq) + 7 H2O(l) + 14 OH-(aq) 4. Allow H+ and OH- to "react" to form H2O if they are on the same side of a half-reaction. H2O2(aq) + 2 OH-(aq) → O2(g) + 2 H2O(l) Cr2O72−(aq) + 14 H2O(l) → 2 Cr3+(aq) + 7 H2O(l) + 14 OH-(aq) 5. Cancel H2O molecules if they appear on opposite sides of the SAME half-reaction. H2O2(aq) + 2 OH-(aq) → O2(g) + 2 H2O(l) Cr2O72−(aq) + 7 H2O(l) → 2 Cr3+(aq) + 14 OH-(aq) 6. Add electrons to either side balance charge. H2O2(aq) + 2 OH-(aq) → O2(g) + 2 H2O(l) + 2 e− Cr2O72−(aq) + 7 H2O(l) + 6 e− → 2 Cr3+(aq) + 14 OH-(aq) 7. Multiply the half-reactions so the number of electrons is equal. 3 H2O2(aq) + 6 OH-(aq) → 3 O2(g) + 6 H2O(l) + 6 e− Cr2O72−(aq) + 7 H2O(l) + 6 e−→ 2 Cr3+(aq) + 14 OH-(aq) 8. Add the half-reactions together and cancel any species that are present on both sides of the reaction. 3 H2O2(aq) + Cr2O72−(aq) + H2O(l) → 3 O2(g) + 2 Cr3+(aq) + 8 OH−(aq)

What is the coefficient for OH−(aq) when SO32−(aq) + MnO4−(aq) → SO42−(aq) + Mn2+(aq) is balanced in basic aqueous solution?

The coefficient for OH−(aq) when SO32−(aq) + MnO4−(aq) → SO42−(aq) + Mn2+(aq) is balanced in basic aqueous solution is 6. 1. Write the two unbalanced half-reactions for the equation. SO32−(aq) → SO42−(aq) MnO4−(aq) → Mn2+(aq) 2. Balance the two half-reactions ((a) balance S and Mn , (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen). SO32−(aq) + H2O(l) → SO42−(aq) + 2 H+(aq) MnO4−(aq) + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) 3. Adjust the two half-reactions for basic solution by adding the enough OH- to both sides of each equation to remove all H+. SO32−(aq) + H2O(l) + 2 OH-(aq) → SO42−(aq) + 2 H+(aq) + 2 OH-(aq) MnO4−(aq) + 8 H+(aq) + 8 OH-(aq) → Mn2+(aq) + 4 H2O(l) + 8 OH-(aq) 4. Allow H+ and OH- to "react" to form H2O if they are on the same side of a half-reaction. SO32−(aq) + H2O(l) + 2 OH-(aq) → SO42−(aq) + 2 H2O(l) MnO4−(aq) + 8 H2O(l) + 8 OH-(aq) → Mn2+(aq) + 4 H2O(l) 5. Cancel H2O molecules if they appear on opposite sides of the SAME half-reaction. SO32−(aq) + 2 OH-(aq) → SO42−(aq) + H2O(l) MnO4−(aq) + 4 H2O(l) → Mn2+(aq) + 8 OH-(aq) 6. Add electrons to either side balance charge. SO32−(aq) + 2 OH-(aq) → SO42−(aq) + H2O(l) + 2 e− MnO4−(aq) + 4 H2O(l) + 5 e− → Mn2+(aq) + 8 OH-(aq) 7. Multiply the half-reactions so the number of electrons is equal. 5 SO32−(aq) + 10 OH-(aq) → 5 SO42−(aq) + 5 H2O(l) + 10 e− 2 MnO4−(aq) + 8 H2O(l) + 10 e− → 2 Mn2+(aq) + 16 OH-(aq) 8. Add the half-reactions together and cancel any species that are present on both sides of the reaction. 5 SO32−(aq) + 2 MnO4−(aq) + 3 H2O(l) → 5 SO42−(aq) + 2 Mn2+(aq) + 6 OH−(aq)

What is the coefficient for OH−(aq) when SO42−(aq)+ Br2(l) → S2O32−(aq) + BrO3−(aq) is balanced in basic aqueous solution?

The coefficient for OH−(aq) when SO42−(aq) + Br2(l) → S2O32-(aq) + BrO3−(aq) is balanced in basic aqueous solution is 2. 1. Write the two unbalanced half-reactions for the equation. SO42−(aq) → S2O32−(aq) Br2(l) → BrO3−(aq) 2. Balance the two half-reactions ((a)balance S and Br, (b) add H2O to balance oxygen, (c) add H+ to balance hydrogen). 2 SO42−(aq) + 10 H+(aq) → S2O32−(aq) + 5 H2O(l) Br2(l) + 6 H2O(l) → 2 BrO3−(aq) + 12 H+(aq) 3. Adjust the two half-reactions for basic solution by adding the enough OH- to both sides of each equation to remove all H+. 2 SO42−(aq) + 10 H+(aq) + 10 OH-(aq) → S2O32−(aq) + 5 H2O(l) + 10 OH-(aq) Br2(l) + 6 H2O(l) + 12 OH-(aq) → 2 BrO3−(aq) + 12 H+(aq) + 12 OH-(aq) 4. Allow H+ and OH- to "react" to form H2O if they are on the same side of a half-reaction. 2 SO42−(aq) + 10 H2O(l) → S2O32−(aq) + 5 H2O(l) + 10 OH-(aq) Br2(l) + 6 H2O(l) + 12 OH-(aq) → 2 BrO3−(aq) + 12 H2O(l) 5. Cancel H2O molecules if they appear on opposite sides of the SAME half-reaction. 2 SO42−(aq) + 5 H2O(l) → S2O32−(aq) + 10 OH-(aq) Br2(l) + 12 OH-(aq) → 2 BrO3−(aq) + 6 H2O(l) 6. Add electrons to either side balance charge. 2 SO42−(aq) + 5 H2O(l) + 8 e− → S2O32−(aq) + 10 OH-(aq) Br2(l) + 12 OH-(aq) → 2 BrO3−(aq) + 6 H2O(l) + 10 e− 7. Multiply the half-reactions so the number of electrons is equal. 10 SO42−(aq) + 25 H2O(l) + 40 e− → 5 S2O32−(aq) + 50 OH-(aq) 4 Br2(l) + 48 OH-(aq) → 8 BrO3−(aq) + 24 H2O(l) + 40 e− 8. Add the half-reactions together and cancel any species that are present on both sides of the reaction. 10 SO42−(aq) + H2O(l) + 4 Br2(l) → 5 S2O32−(aq) + 8 BrO3−(aq) + 2 OH−(aq)

A solution has [Fe2+] of 0.0200 M. What concentration of sulfide is needed before precipitation begins? Ksp of FeS = 3.72 × 10-10

The concentration of sulfide needed to begin precipitation is 1.86 × 10-8 M. The equation for the dissolution of FeS is FeS(s) → Fe2+(aq) + S2−(aq), which gives the solubility constant expression Ksp = [Fe2+][S2−] Substitute the given values and solve for [S2−]. 3.72 × 10-10 = [0.0200][S2−] 1.86 × 10-8 = [S2−]

What mass of CH3COONa must be added to create 300.0 mL of a buffer solution with a pH = 5.00 and an CH3COOH concentration of 0.100 M? The pKa of acetic acid is 4.75.

The mass of CH3COONa needed is 4.4 g.

What is the molar solubility of AgCl if the Ksp is 1.8 × 10-10? AgCl(s) ⇌ Ag+(aq) + Cl−(aq)

The molar solubility of AgCl is 1.3 × 10-5 M. AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Ksp = [Ag+][Cl−] [x][x] = 1.8 × 10-10 x2 = 1.8 × 10-10 x = 1.3 × 10-5 The molar solubility of AgCl is equal to the concentration of Cl− in the solution.

What is the molar solubility of Cr(OH)3 if the Ksp is 1.6 × 10-30? Cr(OH)3(s) ⇌ Cr3+(aq) + 3 OH−(aq)

The molar solubility of Cr(OH)3 is 1.6 x 10-8 M Cr(OH)3(s) ⇌ Cr3+(aq) + 3 OH−(aq) Ksp = [Cr3+][OH−]3 [x][3x]3 = 1.6 × 10-30 27x4 = 1.6 × 10-30 x = 1.6 x 10-8 M The molar solubility of Cr(OH)3 is equal to the concentration of Cr3+, which is 1.6 x 10-8 M.

What is the molar solubility of NiS if the Ksp is 3.0 × 10-20? NiS(s) ⇌ Ni2+(aq) + S2−(aq)

The molar solubility of NiS is 1.7 × 10-10 M. NiS(s) ⇌ Ni2+(aq) + S2−(aq) Ksp = [Ni2+][S2−] [x][x] = 3.0 × 10-20 x2 = 3.0 × 10-20 x = 1.7 × 10-10 The molar solubility of NiS is equal to the concentration of either Ni2+ or S2−.

What is the oxidation state of carbon in Al2(CO3)3?

The oxidation state of carbon in Al2(CO3)3 is +4. The oxidation state of aluminum is +3, which is the same as its charge in an ionic compound. The oxidation state of oxygen is −2. The oxidation state of carbon must be calculated. Assign x as the oxidation state of carbon, and remember the sum of oxidation states for a neutral compound is zero. Therefore, (2 Al)(+3) + (3 C)(x) + (9 O)(−2) = 0 6 + 3x + (−18) = 0 3x + (−12) = 0 3x = 12 x = 4

What is the oxidation state of chromium in K2Cr2O7?

The oxidation state of chromium in K2Cr2O7 is +6. The oxidation state of potassium is +1, which is the same as its charge in an ionic compound. The oxidation state of oxygen is −2. The oxidation state of chromium must be calculated. Assign x as the oxidation state of chromium, and remember the sum of oxidation states for a neutral compound is zero. Therefore, (2 K)(+1) + (2 Cr)(x) + (7 O)(−2) = 0 2 + 2x + (−14) = 0 2x + (−12) = 0 2x = 12 x = 6

What is the oxidation state of sulfur in HSO3−?

The oxidation state of sulfur in HSO3− is +4. The oxidation state of hydrogen is +1, unless it is paired with a metal. The oxidation state of oxygen is −2. The oxidation state of sulfur must be calculated. Assign x as the oxidation state of sulfur, and remember the sum of oxidation states for a polyatomic ion is the charge on the ion. Therefore, (1 H)(+1) + (1 S)(x) + (3 O)(−2) = −1 1 + x + (−6) = −1 x + (−5) = −1 x = 4

A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 11.2 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10−4

The pH after 11.2 mL of NaOH has been added is 3.78. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCOOH + NaOH → HCOONa + H2O I 3.75 × 10−3 mol 1.96 × 10−3 mol 0 ~ C −1.96 × 10−3 mol −1.96 × 10−3 mol +1.96 × 10−3 mol ~ F 1.79 × 10−3 mol 0 1.96 × 10−3 mol ~ After the HCOOH and NaOH react, 1.79 × 10−3 moles of HCOOH and 1.96 × 10−3 moles of HCOO− remain in a total volume of 41.20 mL (30.00 mL HCOOH + 11.20 mL NaOH). Since HCOOH and HCOO− are both in solution, the mixture is a buffer so the Henderson-Hasselbalch equation can be used to solve for the pH. pH equals pKa plus log of concentration of A- divided by the concentration of HA. pH equals negative log of 1.8 × 10^-4 plus log of 8.75 x 10^-4 divided by 2.88 × 10^-3. pH equals 3.74 minus 0.517, so pH equals 3.22. 10.22 is incorrect because it is the pOH, not the pH. 3.70 is incorrect because it is the pH when the amounts of acid and base are inverted in the Henderson-Hasselbalch equation. 10.30 is incorrect because it is the pOH when the amounts of acid and base are inverted in the Henderson-Hasselbalch equation

A 20.00 mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 15.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10−5

The pH after 15.00 mL of HCl has been added is 5.16. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCl + NH3 → NH4Cl + H2O I 3.00 × 10−3 mol 3.00 × 10−3 mol 0 ~ C −3.00 × 10−3 mol −3.00 × 10−3 mol +3.00 × 10−3 mol ~ F 0 0 3.00 × 10−3 mol ~ After the NH3 and HCl react, 3.00 × 10−3 moles of NH4+ remain in a total volume of 30.00 mL (15.00 mL HCl + 20.00 mL NH3). Since only NH4+ is in solution, look at the hydrolysis of the cation to determine the pH. Use the moles of NH4+ and the total volume to find the concentration of NH4+ in solution. NH4+ + H2O → NH3 + H3O+ I 8.57 × 10−2 M ~ 0 0 C −x ~ +x +x E 8.57 × 10−2 M − x ~ x x The Kb of NH3 is given but the Ka of NH4+ is needed. Use Kb and Kw to find Ka. Ka × Kb = Kw Ka × 1.8 × 10−5 = 1.0 × 10−14 Ka = 5.6 × 10−10 The Ka can now be used to solve the equilibrium calculation to find [H3O+]. Kb equals concentration of NH3 times concentration of H3O+ divided by concentration of NH4+, 5.6 × 10^-10 equals x squared over 8.57 × 10^-2 minus x Assume the x is much less than 8.57 × 10−2. 5.6 × 10^-10 equals x squared divided by 8.57 × 10^-2, so x equals 6.9 × 10^-6. [H3O+] is equal to x, so it can be used to find pH. pH = −log(6.9 × 10−6) pH = 5.16 8.84 is incorrect because it is the pOH, not the pH. 2.91 is incorrect because the Kb was used instead of the Ka. 5.89 is incorrect because it uses the moles of NH4+ instead of the molarity of NH4+.

A 25.00 mL sample of 0.175 M HCl is being titrated with 0.250 M NaOH. What is the pH after 19.00 mL of NaOH has been added?

The pH after 19.00 mL of NaOH has been added is 11.92. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of HCl and the added moles of NaOH are determined from the concentrations and volumes of each substance. HCl + NaOH → NaCl + H2O I 4.38 × 10−3 mol 4.75 × 10−3 mol 0 ~ C −4.38 × 10−3 mol −4.38 × 10−3 mol +4.38 × 10−3 mol ~ F 0 3.7 × 10−4 4.38 × 10−3 mol ~ After the NaOH and HCl react, 3.7 × 10−4 moles of NaOH remain in a total volume of 44.00 mL (25.00 mL HCl + 19.00 mL NaOH). The concentration of NaOH in solution is found using the molarity equation. molarity equals 3.7 × 10^-4 moles divided by 4.400 × 10^-2 liters equals 0.0084 molar Since NaOH is a strong base, it completely dissociates into Na+ and OH− ions. Therefore, [NaOH] = [OH−]. pOH = −log [OH−] pOH = −log (0.0084) pOH = 2.08 pH + pOH = 14 pH + 2.08 = 14 pH = 11.92 2.08 is incorrect because it is the pOH, not the pH. 1.83 is incorrect because it is the pOH when the molarity doesn't include the combined volume of the two solutions. 12.17 is incorrect because it is the pH when the molarity doesn't include the combined volume of the two solutions.

A 25.00 mL sample of 0.175 M HCl is being titrated with 0.250 M NaOH. What is the pH after 5.00 mL of NaOH has been added?

The pH after 5.00 mL of NaOH has been added is 0.983. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of HCl and the added moles of NaOH are determined from the concentrations and volumes of each substance. HCl + NaOH → NaCl + H2O I 4.38 × 10−3 mol 1.25 × 10−3 mol 0 ~ C −1.25 × 10−3 mol −1.25 × 10−3 mol +1.25 × 10−3 mol ~ F 3.13 × 10−3 mol 0 1.25 × 10−3 mol ~ After the NaOH and HCl react, 3.13 × 10−3 moles of HCl remain in a total volume of 30.00 mL (25.00 mL HCl + 5.00 mL NaOH). The concentration of HCl in solution is found using the molarity equation. molarity equals 3.13 × 10^-3 moles divided by 3.000 × 10^-2 liters equals 0.104 molar Since HCl is a strong acid, it completely dissociates into H3O+ and Cl− ions. Therefore, [HCl] = [H3O+]. pH = −log [H3O+] pH = −log (0.104) pH = 0.983 0.902 is incorrect because it doesn't account for the combined volume of the two solutions. 2.504 is incorrect because it uses the moles of H3O+ instead of the molarity.

A 30.00 mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 5.00 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10−4

The pH after 5.00 mL of NaOH has been added is 3.22. Set up an ICF (initial-change-final) table to determine the moles of each substance in the solution. The initial moles of NH3 and the added moles of HCl are determined from the concentrations and volumes of each substance. HCOOH + NaOH → HCOONa + H2O I 3.75 × 10−3 mol 8.75 × 10−4 mol 0 ~ C −8.75 × 10−4 mol −8.75 × 10−4 mol +8.75 × 10−4 mol ~ F 2.88 × 10−3 mol 0 8.75 × 10−4 mol ~ After the HCOOH and NaOH react, 2.88 × 10−3 moles of HCOOH and 8.75 × 10−4 moles of HCOO− remain in a total volume of 35.00 mL (30.00 mL HCOOH + 5.00 mL NaOH). Since HCOOH and HCOO− are both in solution, the mixture is a buffer, so the Henderson-Hasselbalch equation can be used to solve for the pH. pH equals pKa plus log of concentration of A- divided by the concentration of HA. pH equals negative log of 1.8 × 10^-4 plus log of 8.75 x 10^-4 divided by 2.88 × 10^-3. pH equals 3.74 minus 0.517, so pH equals 3.22. 10.78 is incorrect because it is the pOH, not the pH. 4.26 is incorrect because it is the pH when the amounts of acid and base are inverted in the Henderson-Hasselbalch equation. 9.74 is incorrect because it is the pOH when the amounts of acid and base are inverted in the Henderson-Hasselbalch equation.

What is the pH of a buffer system that has 0.15 M H2CO3 and 0.20 M HCO3−? The pKa of carbonic acid is 6.37.

The pH of a buffer system that has 0.15 M H2CO3 and 0.20 M HCO3− is 6.49.

What is the pH of a buffer system that has 0.32 M NH4+ and 0.27 M NH3? The pKa of ammonium ion is 9.25.

The pH of a buffer system that has 0.32 M NH4+ and 0.27 M NH3 is 9.18. Use the Henderson-Hasselbach equation to solve for the pH. The pKa value is given (9.25). The acid (HA) is NH4+ and the conjugate base (A−) is NH3. pH equals pKa plus log of the result of the concentration of A− divided by concentration of HA pH equals 9.25 plus log of the result of 0.27 divided by 0.32, which equals 9.25 minus 0.07 which is 9.18 The answer 9.32 is incorrect since this would result from putting the incorrect ratio (0.32/0.27) instead of (0.27/0.32) into the equation. The answer 10.09 is incorrect since this would result from not taking the log of the ratio. The answer 10.44 is incorrect since this would result from putting the incorrect ratio (0.32/0.27) instead of (0.27/0.32) into the equation and not taking the log of the ratio.

Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?

The pair that can create a buffer solution is HF and NaF. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, NaF will completely dissociate to form Na+ and F−. Therefore, HF and F− are a conjugate acid-base pair. HCl and H2O cannot form a buffer because HCl is a strong acid. Additionally, the two substances are not conjugates of one another. HCl and NaOH cannot form a buffer because HCl is a strong acid and NaOH is a strong base. Additionally, the two substances are not conjugates of one another. H3PO4 and Na3PO4 cannot form a buffer because they are not conjugates of one another. H3PO4 is a weak acid but a conjugate acid-base pair differ by a single proton, H+.

Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?

The pair that can create a buffer solution is NH3 and NH4Cl. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, NH4Cl will completely dissociate to form NH4+ and Cl−. Therefore, NH3 and NH4Cl are a conjugate acid-base pair. CH3CO2H and HCN cannot form a buffer because both are weak acids and therefore not a conjugate acid-base pair. NaCl and H2O cannot form a buffer because NaCl is a neutral salt and water is not a conjugate of either Na+ or Cl−. H2SO4 and Na2SO4 cannot form a buffer because they are not conjugates of one another. H2SO4 is a strong acid and a conjugate acid-base pair differ by a single proton, H+.

Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?

The pair that can create a buffer solution is NaH2PO4 and Na2HPO4. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, both compounds will completely dissociate and result in H2PO4− and HPO42− ions, which differ by a single proton, H+. Therefore, these two compounds will form a conjugate acid-base pair. NaOH and H2O cannot form a buffer because NaOH is a strong base. Additionally, the two substances are not conjugates of one another. NaCl and KCl cannot form a buffer because they are both neutral salts and are not a conjugate acid-base pair. NH3 and H2O cannot form a buffer because while NH3 is a weak base, H2O is not its conjugate acid. previous

Which pair of solutions will form an effective buffer?

The pair that can create an effective buffer solution is 0.40 M HF and 0.30 M NaF. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An effective buffer has an acid:base ratio between 1:10 and 10:1. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, NaF will completely dissociate to form Na+ and F−. The acid:base ratio is 1.33:1 so it is in the range of concentrations for an effective buffer. 0.10 M HF and 0.75 MgF2 could form a buffer because HF and F− are conjugate acid-base pair. However, the acid:base ratio is 1:15 because the concentration of F− is 1.5 M because there are two F− ions for each unit of MgF2. The concentration ratio is outside the effective range. 0.40 M HCl and 0.60 M NaOH are not a conjugate acid-base pair so it cannot act as a buffer. H2SO4 and Na2SO4 cannot form a buffer because they are not conjugates of one another because a conjugate acid-base pair differ by a single proton, H+. Also, H2SO4 is a strong acid.

Which pair of solutions will form an effective buffer? 0.40 M HF and 0.30 M NaF Unselected 0.40 M HCl and 0.60 M NaOH Unselected 0.25 M H2SO4 and 0.25 M Na2SO4 Unselected 0.10 M HF and 0.75 MgF2 Unselected I DON'T KNOW YET Most buffers are a mixture of a weak acid and its conjugate base. Buffers resist changes in pH when small amounts of acid or base are added. The effective pH range of a buffer depends on the pKa of the acid and the relative concentrations of the acid, HA, and its conjugate base, A−. A buffer is effective when the ratio of the acid:base concentrations is between 10:1 and 1:10. Ratios outside of this range are not effective as buffers. Which pair of solutions will form an effective buffer?

Which pair of solutions will form an effective buffer? 0.75 M H3PO4 and 0.45 M NaH2PO4 Unselected 1.0 M H2SO4 and 1.25 M NaHSO4 Unselected 0.50 M NH3 and 0.50 M HCl Unselected 0.80 M CH3COOH and 0.75 M HCl

The pair that can create an effective buffer solution is 0.75 M H3PO4 and 0.45 M NaH2PO4. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An effective buffer has an acid:base ratio between 1:10 and 10:1. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, NaH2PO4 will completely dissociate to form Na+ and H2PO4−. H3PO4 and H2PO4− differ by a single proton. The acid:base ratio is 1.7:1 so it is in the range of concentrations for an effective buffer. 1.0 M H2SO4 and 1.25 M NaHSO4 are conjugates of one another but H2SO4 is a strong acid and cannot be part of a buffer with its conjugate. 0.50 M NH3 and 0.50 M HCl are not a conjugate acid-base pair and will not lead to the formation of a buffer regardless of the acid:base ratio of the concentrations. 0.80 M CH3COOH and 0.75 M HCl are both acids and will not form a buffer solution because they are not conjugates of one another.

Which pair of solutions will form an effective buffer? 1.0 M HCN and 0.30 M KCN Unselected 0.50 M NH3 and 0.50 M CH3COOH Unselected 2.0 M CH3COOH and 0.10 M CH3COONa Unselected 0.35 M HF and 0.45 M HNO2

The pair that can create an effective buffer solution is 1.0 M HCN and 0.30 M KCN. A buffer is composed of a weak acid and its conjugate base or a weak base and its conjugate acid. An effective buffer has an acid:base ratio between 1:10 and 10:1. An acid or base differs from its conjugate by a single proton, H+. In aqueous solution, KCN will completely dissociate to form K+ and CN−. The acid:base ratio is 1:0.3 so it is in the range of concentrations for an effective buffer. 0.50 M NH3 and 0.50 M CH3COOH are not a conjugate acid base pair and will not lead to the formation of a buffer regardless of the acid:base ratio of the concentrations. 2.0 M CH3COOH and 0.10 M CH3COONa are a conjugate acid-base pair because CH3COONa completely dissociates into CH3COO− and Na+. However, the acid:base ratio is outside the range of an effective buffer. 0.35 M HF and 0.45 M HNO2 are both weak acids and cannot form a buffer because they are not conjugates of one another.

The equilibrium constant for the given reaction is 1.2 × 103 at 668 K. After 45 seconds, [CO] = 0.045 M, [Cl2] = 0.026 M, and [COCl2] = 1.4 M. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? CO(g) + Cl2(g) ⇌ COCl2(g)

The reaction is at equilibrium because Q = K. Q= [COCl2] /[CO][Cl2] Q= [1.4] / [0.045][0.026] Q= 1.2x10^3 Q is equal to K, so the reaction is at equilibrium.

The equilibrium constant for the given reaction is 14.5 at 483 K. After 30 seconds, [CO] = 0.25 M, [H2] = 0.30 M, and [CH3OH] = 0.75. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? CO(g) + 2 H2(g) ⇌ CH3OH(g)

The reaction will proceed to the left because Q > K. Q= [CH3OH] / [CO][H2]2 Q= [0.75] / [0.25][0.30]2 Q= 33 Q is greater than K, so the reaction will proceed to the left. As the concentration of the product decreases and the concentrations of the reactants increase, the value of Q will decrease until it reaches the value of K.

Which substance is the reducing agent in the following reaction? CO2(g) + H2(g) → 2 CO(g) + H2O(g)

The reducing agent in the reaction is H2(g). To determine the oxidized species, reduced species, oxidizing agent, and reducing agent, determine the oxidation states of each element in the reaction. CO2(g) + H2(g) → 2 CO(g) + H2O(g) The oxidation state of carbon goes from +4 to +2. The oxidation state of hydrogen goes from 0 to +1. The oxidation state of oxygen is −2 and does not change in this reaction. In the process, CO2(g) is reduced to CO(g), and H2(g) is oxidized to H2O(g). The species that is oxidized, H2(g), is the reducing agent, and the species that is reduced, CO2(g), is the oxidizing agent.

Which of these statements about the common-ion effect is most correct? -The solubility of a salt MA is affected equally by addition of either A− or a non-common ion. -The common-ion effect does not apply to unusual ions like SO32−. -The solubility of a salt MA is decreased in a solution that already contains either M+ or A−. -Common ions alter the equilibrium constant for the reaction of an ionic solid with water.-

The solubility of a salt MA is decreased in a solution that already contains either M+ or A−.

Consider the following equilibrium reaction and equilibrium constant: CO(g) + Cl2(g) ⇌ COCl2(g) K = 5.0 Is this equilibrium reaction a product favored equilibrium with mostly products, a reactant favored equilibrium with mostly reactants, or a true equilibrium with both reactants and products?

True equilibrium with both reactants and products WHY? the equilibrium constant is close to approximately 1. K= 5.0

Consider the following equilibrium reaction and equilibrium constant: 2 NH3(g) ⇌ N2(g) + 3 H2(g) K = 9.4 Is this equilibrium reaction a product favored equilibrium with mostly products, a reactant favored equilibrium with mostly reactants, or a true equilibrium with both reactants and products?

True equilibrium with both reactants and products WHY? the equilibrium constant is close to approximately 1. K= 9.4

How many electrons are needed to balance each of the following half-reactions? Reduction half-reaction: Sn2+(aq) + _e− → Sn(s) Oxidation half-reaction: Co(s) → Co2+(aq)+ _e− Overall reaction: Sn2+(aq) + Co(s) → Sn(s) + Co2+(aq)

Two electrons are needed to balance each of the half-reactions. Reduction half-reaction: Sn2+(aq) + 2 e− → Sn(s) Oxidation half-reaction: Co(s) → Co2+(aq) + 2 e− Overall reaction: Sn2+(aq) + Co(s) → Sn(s) + Co2+(aq) Since same number of electrons are present on the left side of the first half-reaction and the right side of the second half-reaction, they cancel out. Notice that both sides of the overall redox reaction have that the same total charge and the same number and types of atoms.

Which one of the following is a nonspontaneous process?

Water decomposing to its elements at 25°C and 1 atm

What can be said about the relative amounts of reactants and products when K is less than one?

When K < 1, there are more reactants than products. The law of mass action includes the concentrations of products over reactants; so when K is less than one, the concentration of products is less than the concentration of reactants.

What can be said about the relative amounts of reactants and products when K is greater than one?

When K > 1, there are more products than reactants. The law of mass action includes the concentrations of products over reactants; so when K is greater than one, the concentration of reactants is less than the concentration of products.

What can be said about the relative amounts of reactants and products when K is approximately one?

When K is approximately 1, the amounts of reactants and products are approximately the same. The law of mass action includes the concentrations of products over reactants. For K to be approximately 1, the concentrations of reactants and products must be approximately the same.

A solution is buffered at a pH of 6.5 and a small amount of base is added. What happens to the pH of the solution?

When base is added to a buffer solution, the pH will increase a small amount. Because the solution is buffered, there will not be a drastic change nor will it remain completely unchanged. Buffers help maintain the pH of a solution but there are variations when acid or base is added.

What is ∆G when ∆G° is −257.2 kJ and the pressure of each gas is 0.0358 atm at 25°C? CO(g) + ½ O2(g) ⇌ CO2(g)

When ∆G° is −257.2 kJ and the pressure of each gas is 0.0358 atm, ∆G equals −253.1 kJ.

What are the values of ∆H and ∆S so that Gibbs free energy, ∆G, is negative at all temperatures?

When ∆H < 0 and ∆S > 0, the value of Gibbs free energy, ∆G, will be negative at all temperatures. If ∆H < 0 and ∆S < 0, then ∆G will be negative at low temperatures and positive at high temperatures. If ∆H > 0 and ∆S > 0, then ∆G will be positive at low temperatures and negative at high temperatures. If ∆H > 0 and ∆S < 0, then ∆G will be positive at all temperatures.

What are the values of ∆H and ∆S so that Gibbs free energy, ∆G, is positive at all temperatures?

When ∆H > 0 and ∆S < 0, the value of Gibbs free energy, ∆G, will be positive at all temperatures. If ∆H < 0 and ∆S < 0, then ∆G will be negative at low temperatures and positive at high temperatures. If ∆H > 0 and ∆S > 0, then ∆G will be positive at low temperatures and negative at high temperatures. If ∆H < 0 and ∆S > 0, then ∆G will be negative at all temperatures.

For a reaction, ∆H and ∆S are both negative. Which choice best describes when the reaction will be spontaneous?

When ∆H and ∆S for a reaction are both negative, it will be spontaneous at low temperatures. For a spontaneous reaction, ∆G < 0. The equation that relates ∆G, ∆H, and ∆S is ∆G = ∆H − T∆S. If ∆H and ∆S are both negative, then ∆G will be negative when the magnitude of the second term (−T∆S) is less than the magnitude of ∆H.

For a reaction, ∆H and ∆S are both positive. Which choice best describes when the reaction will be spontaneous?

When ∆H and ∆S for a reaction are both positive, it will be spontaneous at high temperatures. For a spontaneous reaction, ∆G < 0. The equation that relates ∆G, ∆H, and ∆S is ∆G = ∆H − T∆S. If ∆H and ∆S are both positive, then ∆G will be negative when the second term (T∆S) is greater than ∆H.

For a reaction, ∆H is negative and ∆S is positive. Which choice best describes when the reaction will be spontaneous?

When ∆H is negative and ∆S is positive, it will be spontaneous at all temperatures. For a spontaneous reaction, ∆G < 0. The equation that relates ∆G, ∆H, and ∆S is ∆G = ∆H − T∆S. If ∆H is negative and ∆S is positive, then ∆G will always be negative and the reaction will be spontaneous, regardless of the temperature.

Given ∆Ssystem equals 412 J and ∆Suniverse equals −186 J, what is ∆Ssurroundings?

When ∆Ssystem equals 412 J and ∆Suniverse equals −186 J, ∆Ssurroundings equals −598 J. ∆Suniverse = ∆Ssystem + ∆Ssurroundings -186 J = 412 J + ∆Ssurroundings ∆Ssurroundings = -598 J

Consider the reaction 2XY+Z2⇌2XYZ which has a rate law of rate= k[XY][Z2] Select a possible mechanism for the reaction.

X+Y-> XYZ + Z XY + Z -> XYZ

Which equation represents the integrated rate law for a zero-order reaction?

[A]t = -kt + [A]0

What is the concentration of CO, if the concentration of Cl2 was measured to be 0.75 M, COCl2 was measured to be 2.21 M, and the equilibrium constant, K, is 5.3? CO(g) + Cl2(g) ⇌ COCl2(g)

[CO] = 0.56 M 5.3 = [2.21] / [CO][0.75] [CO] = 0.56 M

What is the concentration of HCl, if the concentration of H2 was measured to be 3.8 × 10-5 M, the concentration of Cl2 was measured to be 4.6 × 10-6 M, and the equilibrium constant, K, is 2.5 × 1034? H2(g) + Cl2(g) ⇌ 2 HCl(g)

[HCl] = 2.1 × 10^12 M

Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 6.56×1010 s ([HI]t) for a reaction starting under the condition in the simulation? Express your answer in moles per liters to three significant figures.

[HI]t = 2.38×10−3 mol/L In this problem, taking into account the initial concentration of HI , you calculated the concentration of HI at time t = 6.56×1010 s . Similarly, if k and the concentration of HI at a particular time are known, the initial concentration of HI can be calculated. 1/[Ht] - 1/[Ho] = kt

What is the concentration of NOBr, if the concentration of NO was measured to be 0.89 M, Br2 was measured to be 0.562 M, and the equilibrium constant, K, is 1.3 × 10-2? 2 NO(g) + Br2(g) ⇌ 2 NOBr(g)

[NOBr] = 0.076 M

What is the concentration of NO, if the concentration of N2 and O2 were measured to be 152 M, and the equilibrium constant, K, is 2.0 × 10-9? N2(g) + O2(g) ⇌ 2 NO(g)

[NO] = 6.8 × 10-3 M 2.0x10^-9 = [NO]2 / [152][152] [NO] = 6.8 x10^-3 M

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Using the rate constant found in Part B, calculate the concentration of sucrose at 140 min if the initial sucrose concentration were 0.316 M and the reaction were zero order in sucrose. Express your answers using two significant figures.

[sucrose]140 = 0 M

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Using the rate constant found in Part B, calculate the concentration of sucrose at 210 min if the initial sucrose concentration were 0.316 M and the reaction were zero order in sucrose. Express your answers using two significant figures.

[sucrose]140 = 0 M

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Using the rate constant found in Part B, calculate the concentration of sucrose at 80 min if the initial sucrose concentration were 0.316 M and the reaction were zero order in sucrose.

[sucrose]80 = 2.2×10−2 M

Give the complete symbol, with superscript and subscript, for a beta particle.

^0_-1e

Give the symbol to complete the balanced nuclear equation.24/11Na-> ?+24/12 Mg

^0_-1e

Identify the balanced nuclear equation for the beta decay of ^10_4Be

^10_4Be->^0_-1e+^10_5B

Identify the balanced nuclear equation for the alpha decay of ^212_84Po

^212_84Po->^4_2He+^208_82Pb

Which option contains the terms that best complete the following sentence? A base _____ a _____ to become an acid.

accepts, proton

For the titration of a strong acid with a strong base, the pH at the half-equivalence point volume is _________.

acidic

For the titration of a weak base with a strong acid, the pH at the equivalence point is _________.

acidic For the titration of a weak base with a strong acid, the pH at the equivalence point is acidic. When a weak base reacts with a strong acid, the resulting salt contains the cation of the weak base and the anion of the strong acid. The anion will not undergo hydrolysis but the cation will. When the cation undergoes hydrolysis, it will produce H3O+ ions in solution. BH+ + H2O ⇌ B + H3O+

In the reaction shown below, which substance is reduced? 3 Na(s) + AlCl3(aq) → 3 NaCl(aq) + Al(s)

aluminum in AlCl3(aq)

For the titration of a weak acid with a strong base, the pH at the equivalence point is _________.

basic For the titration of a weak acid with a strong base, the pH at the equivalence point is basic. When a weak acid reacts with a strong base, the resulting salt contains the cation of the strong base and the anion of the weak acid. The cation will not undergo hydrolysis but the anion will. When the anion undergoes hydrolysis, it will produce OH− ions in solution. A− + H2O ⇌ HA + OH−

The titration of HCOOH with NaOH produces a salt and water. The resulting salt is _________.

basic The titration of HCOOH with NaOH produces a salt and water. The resulting salt is basic because HCOOH is a weak acid and NaOH is a strong base. When a weak acid reacts with a strong base, the resulting salt contains the cation of the strong base and the anion of the weak acid. The cation will not undergo hydrolysis but the anion will. When the anion undergoes hydrolysis, it will produce OH− ions in solution. HCOO− + H2O ⇌ HCOOH + H3O+

In the titration of a weak base with a strong acid with a 1:1 ratio, when does the pOH equal to pKb?

basic For the titration of a weak base with a strong acid, the pH at the half-equivalence point volume is basic. As acid is added to the base, it is neutralized, therefore only base remains in solution at any volume of added acid before the equivalence point volume

In the reaction shown below, which substance is reduced? 3 CuCl2(aq) + 2 Al(s) → 2 AlCl3(aq) + 3 Cu(s)

copper in CuCl2(aq)

Which option contains the terms that best complete the following sentence? An acid _____ a _____ to become a base.

donates, proton

Another first-order reaction also has a rate constant of 2.50×10−2s−1 at 23 ∘C. What is the value of k at 50 ∘C if E2 = 133 kJ/mol ? Express your answer in reciprocal seconds to three significant figures.

k = 2.28 s−1 To calculate the rate constant k2 at some temperature T2 when you know the activation energy and the rate constant k1 at some other temperature T1 , use the two-point form of the Arrhenius equation: lnk1k2=EaR(1T2−1T1) where Ea is the activation energy and R is the gas constant. Since the units of the gas constant are J/(mol⋅K) , begin by converting the activation energy from kJ/mol to J/mol using the appropriate conversion factor: Ea==133kJmol×1000J1kJ1.33×105J/mol Then convert the temperatures from ∘C to K . As shown in Part A, a temperature of 23 ∘C is equal to 296 K, and a temperature of 50 ∘C is equal to 323 K . Now calculate the right side of the two-point form of the Arrhenius equation: lnk1k2==1.33×105Jmol8.314Jmol⋅K(1323K−1296K)−4.51 Then calculate the ratio of the rate constants: k1k2==e−4.511.10×10−2 and rearrange the resulting equation to solve it for k2 : k2===k11.10×10−22.50×10−2s−11.10×10−22.28s−1

Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6. At 23 ∘C and in 0.5 M HCl, the following data were obtained for the disappearance of sucrose: Time (min) C12H22O11(M) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 What is the rate constant? Express your answers using three significant figures.

k = 3.67×10−3 min−1 The first-order rate constant ( k ) is equal to the negative of the slope produced by plotting the natural log of the concentration over time. In order to find the slope of the plot, the natural log of each concentration must first be calculated. [C12H22O11] ln [C12H22O11] 0.316 − 1.15 0.274 − 1.30 0.238 − 1.44 0.190 − 1.66 0.146 − 1.92 These natural log values can then be used to plot a trend line, which will have a slope of −3.67×10−3 . Since k is the negative of the slope, the value of k will be 3.67×10−3 min−1 .

A certain first-order reaction has a rate constant of 2.50×10−2s−1 at 23 ∘C. What is the value of k at 50 ∘C if Ea = 74.5 kJ/mol ? Express your answer in reciprocal seconds to three significant figures.

k =0.313 s−1 FIRST ORDER ln(k1/k2) = Ea/r * (1/T2-1/T1) To calculate the rate constant k2 at some temperature T2 when you know the activation energy and the rate constant k1 at some other temperature T1 , use the two-point form of the Arrhenius equation: lnk1k2=EaR(1T2−1T1) where Ea is the activation energy and R is the gas constant. Since the units of the gas constant are J/(mol⋅K) , begin by converting the activation energy from kJ/mol to J/mol using the appropriate conversion factor: Ea==74.5kJmol×1000J1kJ7.45×104J/mol Then convert the temperatures from ∘C to K : T1(K)T2(K)======T1(∘C)+273.1523∘C+273.15296KT2(∘C)+273.1550∘C+273.15323K Now calculate the right side of the two-point form of the Arrhenius equation: lnk1k2==7.45×104Jmol8.314Jmol⋅K(1323K−1296K)−2.53 Then calculate the ratio of the rate constants: k1k2==e−2.537.98×10−2 and rearrange the resulting equation to solve it for k2 : k2===k17.98×10−22.50×10−2s−17.98×10−20.313s−1

k =2.28 s−1 FIRST ORDER ln(k1/k2) = Ea/r * (1/T2-1/T1) Correct answer is shown. Your answer 2.29 s−1 was either rounded differently or used a different number of significant figures than required for this part. To calculate the rate constant k2 at some temperature T2 when you know the activation energy and the rate constant k1 at some other temperature T1 , use the two-point form of the Arrhenius equation: lnk1k2=EaR(1T2−1T1) where Ea is the activation energy and R is the gas constant. Since the units of the gas constant are J/(mol⋅K) , begin by converting the activation energy from kJ/mol to J/mol using the appropriate conversion factor: Ea==133kJmol×1000J1kJ1.33×105J/mol Then convert the temperatures from ∘C to K . As shown in Part A, a temperature of 23 ∘C is equal to 296 K, and a temperature of 50 ∘C is equal to 323 K . Now calculate the right side of the two-point form of the Arrhenius equation: lnk1k2==1.33×105Jmol8.314Jmol⋅K(1323K−1296K)−4.51 Then calculate the ratio of the rate constants: k1k2==e−4.511.10×10−2 and rearrange the resulting equation to solve it for k2 : k2===k11.10×10−22.50×10−2s−11.10×10−22.28s−1

k =230 M−2⋅s−1 rate=k[ClO2]^2[OH−] rate/ [ClO2]^2[OH-] = k

The gas-phase decomposition of SO2Cl2 is first order in SO2Cl2: SO2Cl2(g)→SO2(g)+Cl2(g) At 600 K the half-life for this process is 2.3×105 s. What is the rate constant at this tterm-47emperature? Express your answer in reciprocal seconds to two significant figures.

k =3.0×10−6 s−1 The expression for the half-life of a first-order reaction is t1/2=0.693k where t1/2 is the half-life for a reaction and k is the rate constant. Rearrange this equation to solve it for k and substitute the given quantity: k===0.693t1/20.6932.3×105s3.0×10−6s−1

A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction Consider the following multistep reaction: A+B→AB(slow) A+AB→A2B(fast)-------------------- 2A+B→A2B(overall) Based on this mechanism, determine the rate law for the overall reaction.

k*[A]*[B]

Consider the following multistep reaction: C+D⇌CD(fast) CD+D→CD2(slow) CD2+D→CD3(fast)--------------------- C+3D→CD3(overall) Based on this mechanism, determine the rate law for the overall reaction.

k*[C]*[D]^2

m =0.957 mg

Which expression best defines the rate of the reaction with respect to O2, given the following reaction? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

rate = -1/5 - (▵ [O2] / ▵t) The equation that best represents the rate of change of O2 for the following reaction is rate equals negative 1 over 5 times change in concentration of oxygen over change in time. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Oxygen is a reactant which is being consumed so the equation must have a negative sign. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is ⅕ of the change in concentration of oxygen over the change in time.

The initial rate of reaction was measured for 2 ClO2(aq) + 2 OH−(aq) → ClO3−(aq) + ClO2−(aq) + H2O(l) at several different concentrations of reactants. What is the rate law for the reaction? Trial [ClO2] (M) [OH-] (M) Initial rate (M/s) 1 0.100 0.100 0.210 2 0.200 0.100 0.840 3 0.200 0.200 1.680

rate = k[ClO2]2[OH−] The rate law for the reaction is rate = k[ClO2]2[OH−] based on the given data. 2 ClO2(aq) + 2 OH−(aq) → ClO3−(aq) + ClO2−(aq) + H2O(l) Trial [ClO2] (M) [OH-] (M) Initial rate (M/s) 1 0.100 0.100 0.210 2 0.200 0.100 0.840 3 0.200 0.200 1.680 Comparing trials 1 and 3, the concentration of ClO2 is doubled while the concentration of OH− is constant. With this change, the initial rate increases by a factor of four which indicates the reaction is second order with respect to ClO2. Comparing trials 2 and 3, the concentration of OH− is doubled while the concentration of ClO2 is constant. With this change, the initial rate doubles which indicates the reaction is first order with respect to OH−.

The initial rate of reaction was measured for F2(g) + 2 ClO2(g) → 2 FClO2(g) at several different concentrations of reactants. What is the rate law for the reaction? Trial [F2] (M) [ClO2] (M) Initial rate (M/s) 1 0.050 0.025 0.130 2 0.050 0.050 0.260 3 0.200 0.025 0.520

rate = k[F2][ClO2] Comparing trials 1 and 3, the concentration of F2 is multiplied by four while the concentration of ClO2 is constant. With this change, the initial rate also increases by a factor of four which indicates the reaction is first order with respect to F2. Comparing trials 1 and 2, the concentration of ClO2 is doubled while the concentration of F2 is constant. With this change, the initial rate doubles which indicates the reaction is first order with respect to ClO2.

The initial rate of reaction was measured for H3PO4(aq) + 3 I−(aq) + 2 H+(aq) → H3PO3(aq) + I3−(aq) + H2O(l) at several different concentrations of reactants. What is the rate law for the reaction? Trial [H3PO4] (M) [I-] (M) [H+] (M) Initial rate (M/s) 1 0.150 0.150 0.150 0.220 2 0.300 0.150 0.150 0.440 3 0.150 0.450 0.150 0.660 4 0.300 0.150 0.300 1.760

rate = k[H3PO4][I−][H+]2 Comparing trials 1 and 2, the concentration of H3PO4 is doubled while the concentrations of the other species are constant. With this change, the initial rate also doubles which indicates the reaction is first order with respect to H3PO4. Comparing trials 1 and 3, the concentration of I− is tripled while the concentrations of the other species are constant. With this change, the initial rate triples which indicates the reaction is first order with respect to I−. Comparing trials 2 and 4, the concentration of H+ is doubled while the concentrations of the other species are constant. With this change, the initial rate increases by a factor of four which indicates the reaction is second order with respect to H+.

The initial rate of reaction was measured for 2 NO(g) + Cl2(g) → 2 NOCl(g) at several different concentrations of reactants. What is the rate law for the reaction? Trial [NO] (M) [Cl2] (M) Initial rate (M/s) 1 0.025 0.030 0.432 2 0.100 0.030 1.728 3 0.100 0.015 0.864

rate = k[NO][Cl2] Comparing trials 1 and 2, the concentration of NO is multiplied by four while the concentration of Cl2 is constant. With this change, the initial rate also increases by a factor of four which indicates the reaction is first order with respect to NO. Comparing trials 2 and 3, the concentration of Cl2 is halved while the concentration of NO is constant. With this change, the initial rate halves which indicates the reaction is first order with respect to Cl2.

The initial rate of reaction was measured for NO(g) + NO2(g) + O2(g) → N2O5(g) at several different concentrations of reactants. What is the rate law for the reaction? Trial [NO] (M) [NO2] (M) [O2] (M) Initial Rate (M/s) 1 0.100 0.050 0.100 0.035 2 0.100 0.100 0.100 0.070 3 0.200 0.050 0.100 0.070 4 0.100 0.050 0.200 0.035

rate = k[NO][NO2] Comparing trials 1 and 2, the concentration of NO2 is doubled while the other species are held constant. With this change, the initial rate doubled which indicates the reaction is first order with respect to NO2. Comparing trials 1 and 3, the concentration of NO is doubled while the other species are held constant. With this change, the initial rate doubles which indicates the reaction is first order with respect to NO. Comparing trials 3 and 4, the concentration of O2 is doubled while the other species are held constant. With this change, the initial rate does not change which indicates the reaction is zero order with respect to O2.

Consider the reaction 2H3PO4→P2O5+3H2O Using the information in the following table, calculate the average rate of formation of P2O5 between 10.0 and 40.0 s. Time (s) 0 10.0 20.0 30.0 40.0 50.0 [P2O5] (M) 0 2.70×10−3 5.70×10−3 7.50×10−3 8.70×10−3 9.30×10−3 Express your answer with the appropriate units.

rate of formation of P2O5=Δ[P2O5]/Δt [(8.70*10^-3 M) - (2.70*10^-3 M) ] / (40 s - 10 s) 0.006 M / 30 s 0.0002 M/s

In the hydrogenation of ethylene using a nickel catalyst, the initial concentration of ethylene is 1.50 mol⋅L−1 and its rate constant (k) is 0.0020 mol⋅L−1⋅s−1 . Determine the rate of reaction if it follows a zero-order reaction mechanism. Express your answer to two significant figures and include the appropriate units.

rate of reaction = 2.0×10−3 mol⋅L−1⋅s−1 Integrated rate law for a zero-order reaction The integrated rate law for a chemical reaction is a relationship between the concentrations of the reactants and time. Consider a single reactant A decomposing into products: A→products. For a zero-order reaction, rate=k Because the rate of the reaction is −Δ[A]/Δt, the above rate law can be written as −Δ[A]Δt=k On integrating this differential equation, the integrated rate law for a zero-order reaction is obtained: [A]t=−kt+[A]0 where [A]t is the concentration at time t, k is the rate constant, and [A]0 is the initial concentration. The variation in concentration with time provides a detailed description of how fast the reaction is occurring.

Which expression best defines the rate of the reaction with respect to C3H8, given the following reaction? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

rate= - ▵[C3H8] / ▵t The equation that best represents the rate of change of C3H8 for the following reaction is rate equals negative change in concentration of C3H8 over change in time. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Propane, C3H8, is a reactant which is being consumed so the equation must have a negative sign. The rate of change with respect to a particular reactant must reflect the stoichiometric coefficient of the substance in the balanced chemical equation. The change in the average reaction rate is equal to the change in concentration of propane over the change in time.

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− This rapid reaction gives the following rate data: [OCl−] (M) [I−] (M) Initial Rate (M/s) 1.5×10−3 1.5×10−3 1.36×10−4 3.0×10−3 1.5×10−3 2.72×10−4 1.5×10−3 3.0×10−3 2.72×10−4 Which among the following is the rate law for this reaction?

rate= k[OCl−][I−] When [OCl−] is doubled while holding [I−] constant (first and second experiments), the initial rate doubles (from 1.36×10−4 M/s to 2.72×10−4 M/s ), so the reaction is first order with respect to the hypochlorite ion. Similarly, when [I−] is doubled while holding [OCl−] constant (first and third experiments), the initial rate is doubled, so the reaction is also first order with respect to iodide.

The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2 The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82×10−3s−1. Which of the following is the rate law for this reaction? What is the rate of reaction when [N2O5]=0.0240 ? Express the rate in molarity per second to three significant figures.

rate=4.82×10−3s−1[N2O5] The value of the rate constant is independent of reactant concentrations. It is stated that the reaction is first order with respect to N2O5 , so the exponent on [N2O5] in the rate law is 1.

rate=k[ClO2]^2[OH−] The coefficients in the balanced chemical equation are used to determine the exponents when you describe an equilibrium relationship, but the exponents in a rate law must be experimentally determined. The order of the reaction depends on the order with respect to each reactant. Experiments 1 and 2 show that the only change in initial concentration was to that of ClO2 , where tripling the concentration produced a nine-fold increase in the rate. Therefore, the reaction is second order with respect to [ClO2] . Experiments 2 and 3 show that the only change in initial concentration was to that of OH− , where tripling the concentration produced a three-fold increase in the rate. Therefore, the reaction is first order with respect to [OH−] .

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically. The reaction of compound A forming compound B was studied and the following data were collected: Time (s) [A] (M) 0. 0.184 200. 0.129 500. 0.069 800. 0.031 1200. 0.019 1500. 0.016 What is the average reaction rate between 0. and 1500. s? Express your answer to three significant figures and include the appropriate units.

rate=−Δ[A]/Δt - (0.016-0.184) / 1500 1.12×10−4 M/s

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically. The reaction of compound A forming compound B was studied and the following data were collected: Time (s) [A] (M) 0. 0.184 200. 0.129 500. 0.069 800. 0.031 1200. 0.019 1500. 0.016 What is the average reaction rate between 200. and 1200. ss ? Express your answer to three significant figures and include the appropriate units.

rate=−Δ[A]/Δt rate = 1.10×10−4 Ms - (0.019-0.129) / (1200 - 200)

The gas-phase decomposition of SO2Cl2 is first order in SO2Cl2: SO2Cl2(g)→SO2(g)+Cl2(g) At 600 K the half-life for this process is 2.3×105 s. At 320 ∘C the rate constant is 2.2×10−5 s−1. What is the half-life at this temperature? Express your answer in seconds to two significant figures.

t1/2 = 3.2×104 s Substitute the given rate constant, k , into the expression for the half-life, t1/2 , of a first-order reaction to calculate the half-life of the gas-phase decomposition of SO2Cl2 at 320 ∘C : t1/2===0.693k0.6932.2×10−5s−13.2×104s

The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which the amount of reactant or its concentration decreases to one-half of its initial value.Determine the half-life for the reaction in Part B using the integrated rate law, given that the initial concentration is 1.50 mol⋅L−1 and the rate constant is 0.0020 mol⋅L−1⋅s−1 . Express your answer to two significant figures and include the appropriate units.

t1/2 = 3.8×102 s The faster the reaction, the shorter the half-life for that reaction. The rate of the reaction is proportional to the rate constant; thus, the larger the rate constant, the shorter the half-life. [A]o/2 = [A]o - kt1/2 = 1.50/(2 x 0.0020)

t1/2 = 63 days

Nuclear decay is a first-order process. Which equation gives the relationship between half-life and the rate constant, k?

t^(1/2) = 0.693/k

The following data were acquired in an iodination experiment involving acetone. All reaction times are in terms of the rate of disappearance of I2. 1Trial Volume of 0.0010 M I2(mL) 2 Volume of 0.050 M HCl(mL) 3 Volume of 1.0 M acetone(mL) 4 Volume of water(mL) 5 Temperature(mL) 6 Reaction time(s) 1 2 3 4 5 6 A 5.0 10.0 10.0 25.0 25.0 130 B 10.0 10.0 10.0 20.0 25.0 249 C 10.0 20.0 10.0 10.0 25.0 128 D 10.0 10.0 20.0 10.0 25.0 131 E 10.0 10.0 10.0 20.0 42.4 38 In the general rate law rate=k[I2]X[H+]Y[CH3COCH3]Z what are the values of X, Y, and Z? Drag each item to the appropriate bin

x=0 y and z = 1 The rate law for this reaction can thus be written as rate=k[H+][CH3COCH3]

If ∆Suniverse and ∆Ssystem are both positive, what do we know about the sign of ∆Ssurroundings?

∆Ssurroundings can be positive or negative, depending on the magnitude of the values or the system and surroundings. If ∆Ssurroundings is positive, then ∆Suniverse = ∆Ssystem + ∆Ssurroundings will be positive, regardless of the magnitude of the values. If ∆Ssurroundings is negative, then ∆Ssystem must be greater than −∆Ssurroundings for ∆Suniverse to be positive.

What is the rate of change of C3H8 if the rate of change of H2O is 0.375 M/s? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

−0.0938 M/s

What is the rate of change of O2 if the rate of change of H2O is 0.200 M/s? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

−0.250 M/s If the rate of change of H2O is 0.200 M/s, then the rate of change of O2 is -0.250 M/s. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) The equation that compares the rate of change of H2O and O2 is the following because both terms are equal to rate. negative 1 over 5 times change in concentration of O2 over change in time equals 1 over 4 times change in concentration of H2O over change in time Insert the given rate of change for H2O to solve for the rate of change of O2.

What is the rate of change of O2 if the rate of change of CO2 is 0.450 M/s? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

−0.750 M/s If the rate of change of CO2 is 0.450 M/s, then the rate of change of O2 is −0.750 M/s. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) The equation that compares the rate of change of CO2 and O2 is the following because both terms are equal to rate. negative 1 over 5 times change in concentration of O2 over change in time equals 1 over 3 times change in concentration of CO2 over change in time Insert the given rate of change for CO2 to solve for the rate of change of O2.

The equilibrium concentrations of A and B are 0.0241 M and 0.0822 M, respectively. What is ΔG° at 25°C? 2 A(g) ⇌ B(g)

−12.3 kJ

What is the standard entropy change for the reaction? 2 NO2(g) ⇌ N2O4(g) Substance ∆S° (J/mol∙K) NO2(g) 240 N2O4(g) 304

−176 J/K

What is the standard entropy change for the reaction? CO(g) + 3 H2(g) → CH4(g) + H2O(g) Substance ∆S° (J/mol∙K) CO(g) 198 H2(g) 131 CH4(g) 186 H2O(g) 189

−216 J/K

What is ∆G when ∆G° is −32.96 kJ and Q equals 42.15 at 25°C? N2(g) + 3 H2(g) ⇌ 2 NH3(g)

−23.69 kJ

What is the change in the free energy, ∆G°, for the reaction? 4 KClO3(s) + 3 H2S(aq) → 4 KCl(s) + 3 H2SO4(aq) Substance ∆Gf° (kJ/mol) KCLO3(g) -296.3 H2S(g) -27.7 KCL(g) -408.5 H2SO4(g) -744.6

−2599.5 kJ

What is the oxidation state of nitrogen in NH4+?

−3

The equilibrium concentrations of A and B are 0.0354 M and 0.121 M, respectively. What is ΔG° at 25°C? A(g) ⇌ B(g)

−3.04 kJ

What is the change in Gibbs free energy, ∆G, at 25°C given ∆H = −92.22 kJ and ∆S = −198.75 J/K? N2(g) + 3 H2(g) → 2 NH3(g)

−32.99 kJ

What is the change in Gibbs free energy, ∆G, at 25°C given ∆H = −57.20 kJ and ∆S = −175.83 J/K? 2 NO2(g) → N2O4(g)

−4.80 kJ

What is the change in the free energy, ∆G°, for the reaction? 2 CO(g) + O2(g) → 2 CO2(g) Substance ∆Gf° (kJ/mol) CO(g) -137.2 CO2(g) -394.4

−514.4 kJ

What is ∆G when ∆G° is −76.0 kJ and the pressure of each gas is 0.0255 atm at 25°C? 4 HCl(g) + O2(g) ⇌ H2O(g) + 2 Cl2(g)

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