Chem Exam Review

What is the pOH of 1.27e-4 M KOH?

-log(1.27E-4)

What is the overall order of a reaction if its rate law is Rate = k[H2Se2O3][H+]2[I-]3?

1+2+3=6

Using the Boltzmanns formula calculate the entropy of 2.72e14 molecules of a compound if they are randomly distributed in one of 3 ways.

1.38E-23*2.72E14*ln(3)

By how much would the rate of reaction (see below) change if the concentration of H+ increased by a factor of 1.842. The rate law of this reaction is Rate = k[BrO3-][Br-][H+]2. BrO3- + 5Br- + 6H+ ↔ 3Br2 + 3H2O

1.842^2 times the normal rate

What is the pH of 1.93e-3 M KOH?

14-(-log(1.93E-3))

Lysine Formula Mass

146.19 g/mol

aspirin formula mass

176.6 g/mol

glucose formula mass

180.156 g/mol

citric acid formula mass

192.124

What is the rate of consumption of B (M/s), if the rate of consumption of C is 0.265 M/s? Use the reaction below: 3A + 2B + 5C ↔ products

2B = 5C B = 2/5(.265) = .106 M/s

sucrose formula mass

342.3 g/mol

What is the equilibrium concentration of H3O+ in 0.1194 M HNO2 ? Ka = 4.600x10-4

4.600E-4=x^2/(.1194-x), x=H3O+

What is the temperature at which the reaction below becomes spontaneous, if the standard entropies of reactants and products at 25oC are So (C(gr)) = 5.74 J/mol*K, So(H2(g)) = 130.68 J/mol*K, and So (C7H16(g)) = 426.3000J/mol*K and the heat of formation of C7H16(g)) is -187.9000 kJ/mol? 7C(gr) + 8H2(g) ↔ C7H16(g)

7C(gr) + 8H2(g) <----> C7H16(g) DS0rxn = (1*S0f,C7H16)-(7*s0F,C(gr) + 8*S0f,H2(g)) = (1*426.3)-(7*5.74 + 8*130.68) = -659.32 j/mol.k DH0rxn = (1*DH0f,C7H16)-(7*DH0F,C(gr) + 8*DH0f,H2(g)) = (1*-187.9)-(7*0+8*0) = -187.9 kj/mol DG0 = DH0 - TDS0 0 = (-187.9) - (T*-659.32*10^-3) T = 284.99 k

What is a catalyst and how does it work?

A catalyst changes the rate of reaction but is not used up. It increases rate of reaction by providing a different pathway for the reaction that has a lower activation energy.

What is the free energy change at 25oC for the reaction below, if the standard entropies of reactants and products are So (C(gr)) = 5.74 J/mol*K, So(H2(g)) = 130.68 J/mol*K, and So (C5H12(g)) = 349.0000J/mol*K and the heat of formation of C5H12(g) is -146.4400 kJ/mol. C5H12(g) ↔ 5C(gr) + 6H2(g)

C5H12(g) ↔ 5C(gr) + 6H2(g) Entropy of Reaction = 5 * Entropy of C(gr) + 6 * Entropy of H2(g) - Entropy of C5H12(g) => 5 * 5.74 + 6 * 130.68 - 349 => 463.78 J/molK Free energy Change Delta G = Delta H - T Delta S => -146.4400 - (298)/1000 * 463.78 => -284.64644 kJ/mol

Initially there were equilibrium concentrations of 0.3542 M A, 0.2392 M B, and 0.1368 M C. Which change will cause the equilibrium to shift from left to right if the reaction is exothermic? Use the reaction below. 2A(aq) ↔ 3B(aq) +3C(aq) K = 0.990

Decrease the temperature

KF has a face-centered cubic unit cell in which the F- anions occupy corners and face centers, while the cations fit into the hole between adjacent anions. What is its density (in g/cm3) if the ionic radii of K+, F- ions are 132.9 pm and 136.0 pm, respectively? Hint: notice that the cation and anion touch each other and together make up the side of the cube.

Density = ZM/a^3Na Z = no of atoms present in a unit cell = 8*1/8+6*1/2 = 4 M = formula mass = 58.097 g/mol a = edge length = 2.689 *10^-8 cm Na = 6.023*10^23 Density = 4* 58.097/(2.689 *10^-8 )^3*6.023*10^23 = 19.844 g/cc or g/ml

What is the [H3O+] (molarity) of a 0.1230 M methylamine? See reaction below. Kw = Ka x Kb Kb = 3.70e-4

E-14/3.7E-4=x^2/(.1230-x)

From the reaction below, calculate the free energy of formation for C5H8(g) at 25°C, if ΔGfo(CH4(g)) is -50.790 kJ/mol. 5CH4(g) ↔ C5H8(g) + 6H2(g) ΔGo = 464.021 kJ/mol.

G = G-C5H8 + 6* Gf-h2 - (5G-CH4) since Gf-h2 = 0 G = G-C5H8 - (5G-CH4) 464.021 = G-C5H8 + - (5-50.79) G = 464.021 + 5*(-50.79) = 210.071 kJ

How much energy in MeV is released to form one atom of 45Sc from protons, electrons, and neutrons if the nucleus has a mass of 44.955910 amu? The masses of the proton, electron and neutron are 1.00728 amu, 0.000549 amu and 1.00867 amu, respectively. 1.602x10-13 J = 1 MeV; 1 amu = 1.66054x10-27 kg

Givem mass of Sc=44.955910 amu Number of protons and electrons =21 Number of neutrons =24 so Total mass=21*1.00728+21*0.000549+24*1.00867=45.372489 amu so Binding Energy=((Total mass-Given mass)*c2 ) *1.66054*10(-27) =(6.225714*10(-11))/1.602*10-13 = 3.8862*100 Binding Energy =388.62 Mev

Calculate the pOH in 0.2965 M H2S. The stepwise dissociation constants are Ka1 = 1.00e-7 and Ka2 = 1.10e-19

H2S -----------> H^+ (aq) + HS^- (aq) I 0.2965 0 0 C -x +x +x E 0.2965-x +x +x Ka1 = [H^+][HS^-]/[H2S] 1*10^-7 = x*x/0.2965-x 1.*10^-7 *(0.2965-x) = x^2 x = 0.000172 [H^+] = x = 0.000172M PH = -log[H^+] = -log0.000172 = 3.7644 POH = 14-PH = 14-3.7644 = 10.2356

Calculate the [OH-] in 0.2498 M H3PO4. The stepwise dissociation constants are Ka1 = 7.50e-3 and Ka2 = 6.20e-8

H3PO4 <==> H2PO4- + H+ let x amount has reacted Ka1 = 7.50 x 10^-3 = x^2/0.2498 x = [H+] = 0.0433 M Thus, [OH-] = 1 x 10^-14/0.0433 = 2.31 x 10^-13 M

What is the equilibrium molar concentration of Hg2+ in a solution prepared by adding 0.1772 mol of Hg(NO3)2 to a solution of NH3? The final volume of the solution was 1.00 L and the concentration of NH3 was 2.0250 M before equilibrium was established. Hg2+(aq) + 4NH3(aq)↔[Hg(NH3)4]2+(aq) Kf = 1.80e19

Hg2+(aq) + 4NH3(aq)↔[Hg(NH3)4]2+(aq) Kf = 1.8*1019 Hg2+(aq) + 4NH3(aq)↔[Hg(NH3)4]2+(aq) I 0.1772 2.0250 0 C -0.1772 - 4*0.1772 0.1772 E x 1.3162 0.1772 Kf = [ [Hg(NH3)4]2+]/[Hg2+][NH3]^4 1.8E19 = 0.1772/x*(1.3162)^4 x = 0.1772/1.8E19 *(1.3162)^4 = 3.28E-21 [Hg2+] = 3.28*10-21 M

The reaction below is an endothermic reaction. What would one have to do to increase the size of the equilibrium constant?

Increase the temperature

Determine the molar solubility of AgI (Ksp = 8.50e-17) in 2.577 M S2O32- if the complex ion [Ag(S2O3)2]3- forms with a Kf = 2.90e13.

K = Kf × Ksp = 2.90E13 × 2.91E-8 = 8.45 × 105 x² = K × [S203⁻]² = (4.5 × 10⁴)(2.577- 2x)² x = 0.09193/0.11637 = [Ag(S2O3)2]3- = molar solubility

Which equilibrium constant indicates that the reaction will produce mostly products? Some Products? very little products?

K>>>>1 K approx. 1 K<<<<1

What is the conjugate base and base dissociation constant of HCO-3 if its acid dissociation constant is 4.800x10-11 ?

Ka (acid) x Kb (conjugate base) = Kw= E-14 Ka = 4.8E-11 (given) Kb (CO32-) = E-14 /4.8E-11= 0.20E-3 Hence, the dissociation constant of conjugate base of HCO3- is: Kb (CO32-) = 0.20 x 10-3

A mixture containing an initial concentration of 0.1040 M for H2 and 0.0871 M for Br2 is allowed to come to equilibrium (see reaction below). What must be the equilibrium concentration of HBr? Br2(g) ↔ 2HBr(g) Kc = 1.42e3

Kc = [HBr]2 / [H2] [Br2] 1.42 x 10-3 = [HBr]2 / [0.1040-x] [0.0871-x] [HBr]2 = [0.1040-x] [0.0871-x] 1.42 x 10-3 HBr = 5.15 x 10-2 M

A mixture containing an initial concentration of 0.1391 M for both CO and Cl2 is allowed to come to equilibrium (see reaction below). What must be the equilibrium concentration of COCl2? CO(g) + Cl2(g) ↔ COCl2(g) Kc = 1.23e3

Kc = x / (0.1391-x)^2 1.23 x 10^3 = x / (0.1391-x)^2

What is the coefficient, b, in the reaction below, if Kc = 4.69e-6 and Kp = 1.08e-7 at 258.74

Kp=Kc (RT)?n 1.08*10-7=(4.69*10^-6)(0.0821*531.74)^(2-b) 43.52=43.65^(b-2) b-2=1 b=3

A 0.083 M solution of a monoproticacid is known to be 1.07 % ionized. a) what is the PH of the solution? b) calculate the Ka for this acid

Let HA is the monoprotic acid . HA <=> H+ + A- Ka = (x)(x) / ( 0.083 - x) 1.07% = [H+ ] ionised * 100 / 0.083M [H+ ] ionised = 0.083 M * 1.07 /100 = 0.0008881 M = x a) pH = - log [H+ ] = - log ( 0.0008881 ) = 3.0 b ) Ka = ( 0.0008881 )( 0.0008881 ) / ( 0.083-0.0008881) = 9.6E-6

How much energy (in kJ) is required to heat 10.40 g of Ru(s) from 2310.0 K (its melting point) to 3261.4 K. The heat of fusion of Ru is 26.00 kJ/mol and the molar heat capacity of liquid Ru is 24.05J/mol*K.

Molar mass of Ru = 101.07 g/mole Thus, moles of Ru in 10.4g of it = mass/molar mass = 10.4/101.07 = 0.103 Total heat required = Heat required to melt Ru + heat required to bring it to 3261.4 K Thus, total heat = 0.103*26 + 0.103*(24.05/1000)*(3261.4-2310) = 5.028 KJ

For a galvanic cell, the anode has a ____ sign and is the site of ____.

Negative, oxidation

By how much would the rate of reaction (see below) change if the concentration of HCOOH increased by a factor of 2.402. The rate law of this reaction is Rate = k[Br2]. HCOOH(aq) + Br2(aq) ↔ 2Br-(aq) + CO2(g)

No Impact

Which of these nuclides is most likely to be radioactive?

No of proton(p) = atomic number No of neutron (n) = mass no - no of proton. If n/p ratio is greater than 1.5 then element will be radioactive. For Am number of protin =95 Number of neutron=243-95. = 148 So n/p = 148/95 =1.557 which is greater than 1.5 .So it is a radioactive element.

Write balanced equation for the anode of the following galvanic cell Co(s)|Co2+(aq)||Cu2+(aq)|Cu(s) PART B Write balanced equation for the cathode of the following galvanic cell Co(s)|Co2+(aq)||Cu2+(aq)|Cu(s) PART C Write balanced equation for overall cell reactions of the following galvanic cell Co(s)|Co2+(aq)||Cu2+(aq)|Cu(s)

PART (A): - Co (s) ---------> Co2+ (aq.) + 2 e PART (B) Cu2+ (aq) + 2 e ----------> Cu (s) PART (C) Co (s) + Cu2+ (aq.) ---------------> Co2+ (aq.) + Cu(s)

For a galvanic cell, the cathode has a ____ sign and is the site of ____.

Positive, Reduction

entropy change (delta S)

Products minus reactants

During the expansion of 0.9704 mole of CO2 gas, the pressure decreased from 2.304 atm to 1.0 atm. What was the DS (J/K) of the process?

Q = Delta U + WBSince temperature is constant, Delta U = 0.0 J .Q = WB = Integral [ P dV ] V1----->V2Q = WB = [ ( n ) ( R ) ( T ) ] [ ln ( V2/V1 ) ]Delta S = Integral [ dQrev / T ] = Q / TDelta S = [ ( n ) ( R ) ] [ ln ( V2 / V1 ) ] since temp is constant, P1V1 =P2V2 => V2/V1 = P1/P2 Delta S = (0.9704 mol)* ( 8.314 J / gmol - K ) [ ln ( 2.304 atm / 1 atm ) ] Delta S = 6.733 J / K

Initially there were equilibrium concentrations of 0.1400 M A, 0.3333 M B, and 0.1400 M C. Which change will cause Qc to be greater than Kc? Use the reaction below. 5A ↔ 3 B +3 C Kc = 0.988

Qc for the reaction = [B]3 [C]3 / [A]5 From the above equation, we see that a decrease in [A] will increases Qc an increase in [B] and [C] increases Qc. Therefore, the change that will cause Qc to be greater than Kc is a decrease in [A]

What is the standard free-energy change for the reaction Ka for HCN = 4.0x10-10. R = 8.314 J/(mol*K)

Remember to calculate Kb!

Given the following values of equilibrium constants: CuCO3(s) Cu2+(aq) + CO32-(aq) Ksp = 1.4 x 10-10 Cu(NH3)42+ Cu2+(aq) + 4NH3(aq) K = 1.0 x 10-13 a) What is the solubility of CuCO3 in water? b) What is the value of the equilibrium constant for the reaction: CuCO3(s) + 4NH3(aq) Cu(NH3)42+(aq) + CO32-(aq) c) What is the solubility of CuCO3 in 6.00 M NH3?

The computation are expressed below, (a) CuCO3 <--> Cu2+ + CO32- Ksp = [Cu2+][CO32-] = 1.4E-10 s = (Ksp)1/2 = (1.4E-10)1/2 = 1.18E-5 M (b) CuCO3 + 4NH3 <--> Cu(NH3)42+ + CO32- Keq = [Cu(NH3)42+][CO32-] / [CuCO3][NH3]4 given Ksp = 1.0E-13 = [Cu(NH3)42+] [Cu2+] [NH3]4 = 0.00681 M by plugging the values we get, Keq = [0.00681][1.18E-5] / [1.18E-5][0.00681]4 = 3.17E6 (c) The solubility is, CuCO3 + NH3 <--> Cu(NH3)4 + CO3 Ksp =1.0E-13 = [CuCO3][6] [CuCO3] = 1.67E-14 M

The pH of a solution of HClO3 was 3.837. What was the hydronium ion concentration?

[H3O+] = 10-pH 1.46E-4

Initially there were equilibrium concentrations of 0.1204 A, 0.1684 B. What is the answer if the volume is increased by a factor of 2.82? Use the reaction below. 5A(g) ↔ 4 B(g)

concentration = 0.1204 so moles = A = 0.1204 moles (Initially) B = 0.1684 The new concentrations are A = 0.1204 / 2.82 = 0.0426 B = 0.1684 / 2.82 = 0.0597 Kc = [B]^4 / [A]^5 So new Kc = 2.82 times of previous Kc

Gallium metal, ga, crystallizes in a body-centered arrangement. What is the density of Ga if its atomic radius is 147.1 pm?

density = mass/volume volume = s*sqrt3= 4rs = 4r / sqrt(3) s^3= 64/(9*sqrt(3)*r^3 mass = 2M/Na = M= molar mass of gallium = 69.72 and Na = avagadro's numbr = 6.022* 10^-23 mass = 2.31*10^-22 gr volume= 3.912*10^7 pm ^3 density= 5.9*10^-30 gr/pm^3

The bond length of KF is 269.00 pm. What would be the dipole moment, in D, of this compound by assuming a completely ionic bond? Remember e = 1.60x10^-19 C; 1 D = 3.34x10^-30 C*m.

dipole moment= charge on each component * distance between them =1.6*10^-19 * 269*10^-12 =4.304 *10^-29 Cm =4.304 *10^-29 /3.34x10^-30 =12.89 dioptre

What is the change in mass of 127I when it is formed from protons, electrons and neutrons if the atom has a mass of 126.904473 amu? The masses of the proton, electron and neutron are 1.00728 amu, 0.000549 amu and 1.00867 amu, respectively.

iven 127 I we know that atomic number of iodine (I) = 53 so number of protons = number of electrons = 53 now number of nuetrons = mass number - atomic number so number of nuetrons = 127 - 53 number of nuetrons = 74 now total mass = mass of neutrons + mass of protons + mass of electrons so total mass = ( 74 x 1.00867) + ( 53 x 1.00728) + ( 53 x 0.000549) total mass = 128.056517 amu now mass of Iodine = 126.904473 amu so change = mass of iodine finally formed - total mass of particles reacted change = 126.904473 - 128.056517 change = -1.152044 so the change in mass is -1.152044 amu

Radioactive substances decay by first-order kinetics. How many years would be required for a sample containing carbon-14 to decrease to 64.76% of its initial activity? The half-life of carbon-14 is 5.73e3 years.

k = ln2/ 5.73e3 = 1.21*10^-4 A = Ao e^-kt 0.6476A = A e^-(1.21*10^-4t) t = 3590.76 years = 3.59 e 3 years

How much time in seconds would it take for a 2.456 g sample of 137Cs to decay to a mass of 0.9579 g if it has a half-life of 9.4608e+8 s.

k*t0.5=0.693 k=7.32*E-10sec kt=ln(2.456/0.9576) t=ln(2.456/0.9576)/7.32E-10 t=1.28667*E9s

NO2 at 330°C decomposes to form NO and O2 according to second order kinetics (see reaction below). The half-life for this reaction was measured to be 1.55e2 s when the initial concentration of NO2 was 8.37e-3 M. What is the rate constant for this reaction?

k=1/((t1/2)[A0])

The vapor pressure of pure ethanol is greater than the vapor pressure of pure water. Assuming ideal behavior, upon addition of ethanol (EtOH) to water the vapor pressure (vp) of the solution should be

less than the vp of pure EtOH, greater than pure water

The free energy of formation for C10H24(g) is 138.470 kJ/mol. What is the equilibrium constant at 25.0?C for its formation?

lnk= -138.47/(0.00831*298) =-138.47/2.4768 = -55.9163 K= 5.198E-25

What is the cell voltage for the following cell? 3Cd2+(aq) + 2Al(s) <=> 3Cd(s) + 2Al3+(aq) DeltaGo = -729.427 kJ

n = no of electron transfer, 6 F = Faraday constant, 96485C/mol Delta G = -nFE°cell E°cell = - (-729427J/mol/(6 × 96485C/mol) volt = Jule/Coulomb Therefore, E°cell = 1.26V

What is the [H3O+] of a 0.1310 M HOCH2COONa? Answers are expressed in molarity. Ka for HOCH2COOH is 1.32e-4.

pH= 7 +(1/2)(3.847983+log( 0.1310 ))

A buffer made from which one of the acids below would be the best for a ph of 2.288

pKa=-log(H3po4, ka= 7.52e-3)=2.12 Closest to pH

What is the fraction of A remaining after 68.186 minutes if the first order rate constant is 1.129e-4 s^-1

rate constant(k)*time(t) = ln(initial concentration/concentration after time 't') or, (1.129E-4)*68.186*60 = ln(initial concentration/concentration after time 't') or, 0.462 = ln(initial concentration/concentration after time't') or, (initial concentration/concentration after time 't' in seconds) = e0.462 = 1.57 Thus, fraction remaining after 68.186 min = 1/1.57 = 0.63 = 63%

What is the solubility in g/L of N2 at a partial pressure of 283.6 torr, if the solubility at a partial pressure of 417.9 torr is 0.01063 g/L?

solubity of N2 = 417.9*.01063/283.6 = 0.01566 g/L

What is the free energy change for the reaction below, if the reaction quotient, Q, at 25.0oC is equal to 1.04e-7? 5C2H2(g) + 2 H2(g) ↔ C10H14(g) ΔGo = -901.580 kJ.

ΔG=ΔGo+RTlnQ