CHM 1150 Labflow Quizzes Math Problems

What is the correct reading of the temperature in the pictured thermometer? Make sure to report your reading with the appropriate significant figures.

35.9 °C

Sugar is easily soluble in water and has a molar mass of 342.30 g/mol. What is the molar concentration of a 274.6 mL aqueous solution prepared with 73.9 g of sugar?

0.786 M 73.9 g / (342.30 g/mol) = 0.2159 0.2159 / 0.2746 L = 0.786 M

Sugar is easily soluble in water and has a molar mass of 342.30 g/mol. What is the molar concentration of a 228.7 mL aqueous solution prepared with 75.9 g of sugar?

0.97 M 75.9 g / (342.30 g/mol) = 0.2217 0.2217 / 0.2287 L = 0.97 M

Fill in the table by predicting the sign of ΔG and whether or not the process is spontaneous. Reaction conditions Sign of ΔHº Sign of ΔS High T + + Low T + + High T - - Low T - - All T + - All T - +

Sign of ΔGº Spontaneous under standard conditions? - yes + no + no - yes + no - yes Use the equation for Gibbs Free Energy to predict the sign of ΔG and spontaneity: ΔG=ΔH−TΔS Plug in the sign for the variables ΔH and ΔS to predict the sign of ΔG. A negative value for Gibbs Free Energy indicates a spontaneous process, while a positive value will be a non-spontaneous process.

What is the freezing point (in degrees Celcius) of 4.22 kg of water if it contains 194.9 g of CaBr2? The freezing point depression constant for water is 1.86 °C/m and the molar mass of CaBr2 is 199.89 g/mol.

-1.29 °C 194.9 g / 199.89 g/mol = 0.9750 mol 0.9750 mol / 4.22 kg = 0.2310 m ∆T = i x m x K 3 x 0.2310 m x 1.86 °C/m = 1.29 °C 0 - 1.29 °C = -1.29 °C (i = van't Hoff constant = 3 for CaBr2 because it dissociates into 3 ions: Ca2+ and 2Br-)

What is the freezing point (in degrees Celcius) of 3.24 kg of water if it contains 174.6 g of CaBr2? The freezing point depression constant for water is 1.86 °C/m and the molar mass of CaBr2 is 199.89 g/mol.

-1.50 °C 174.6 g / 199.89 g/mol = 0.8735 mol 0.8735 mol / 3.24 kg = 0.2696 m ∆T = i x m x K 3 x 0.2696 m x 1.86 °C/m = 1.50 °C 0 - 1.50 °C = -1.50 °C (i = van't Hoff constant = 3 for CaBr2 because it dissociates into 3 ions: Ca2+ and 2Br-)

Consider the reversible dissolution of lead(II) chloride: PbCl2 (s) ⇌ Pb^2+ (aq) + 2Cl− (aq) Suppose you add 0.2346 g of PbCl2 (s) to 50.0 mL of water. When the solution reaches equilibrium, you find that the concentration of Pb^2+ (aq) is 0.0159 M and the concentration of Cl− (aq) is 0.0318 M. What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2? (Note that the equilibrium constant associated with sparingly soluble salts is called the solubility product constant, Ksp.)

0.0000161 Ksp = [Pb^2+] [Cl−]^2 0.0000161 = [0.0159 M] [0.0318 M]^2

The molar solubility of Ca(OH)2 was experimentally determined to be 0.020 M. Based on this value, what is the Ksp of Ca(OH)2?

0.000032 Ca(OH)2 ⇌ Ca^2+ + 2OH^− Ksp = [Ca^2+] [OH^−]^2 Ksp = (0.020) (2 x 0.020)^2 Ksp = 0.000032

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 20.45 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 of unknown concentration. The endpoint was observed to occur at 14.05 mL. How many moles of KIO3 were titrated?

0.0002045 mol KIO3 = M KIO3 x V KIO3 0.0002045 mol = 0.0100 M x 20.45 mL x (1 L / 1000 mL)

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 25.95 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 of unknown concentration. The endpoint was observed to occur at 12.33 mL. How many moles of KIO3 were titrated?

0.0002595 mol KIO3 = M KIO3 x V KIO3 0.0002595 mol = 0.0100 M x 25.95 mL x (1 L / 1000 mL)

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000523 mol KIO3 were titrated with an unknown solution of Na2S2O3 and the endpoint was reached after 17.80 mL. How many moles of Na2S2O3 did this require?

0.0003138 Redox reaction: IO3^− (aq) + 6S2O3^2− (aq) + 6H+ (aq) ⟶ I− (aq) + 3S4O6^2− (aq) Equation: (moles of KIO3) x (6 mol S2O3^2- / 1 mol IO3^-) = mol S2O3^2- (0.0000523 mol) x (6/1) = 0.0003138 mol S2O3^2-

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.000529 mol KIO3 were titrated with an unknown solution of Na2S2O3 and the endpoint was reached after 17.80 mL. How many moles of Na2S2O3 did this require?

0.003174 Redox reaction: IO3^− (aq) + 6S2O3^2− (aq) + 6H+ (aq) ⟶ I− (aq) + 3S4O6^2− (aq) Equation: (moles of KIO3) x (6 mol S2O3^2- / 1 mol IO3^-) = mol S2O3^2- (0.000529 mol) x (6/1) = 0.003174 mol S2O3^2-

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000157 mol KIO3 were titrated with an unknown solution of Na2S2O3 and the endpoint was reached after 17.64 mL. What is the concentration of the Na2S2O3 solution, in M?

0.00534 Redox reaction: IO3^− (aq) + 6S2O3^2− (aq) + 6H+ (aq) ⟶ I− (aq) + 3S4O6^2− (aq) Equation: (moles of KIO3) x (6 mol S2O3^2- / 1 mol IO3^-) = mol S2O3^2- (0.0000157 mol) x (6/1) = 9.42 x 10^-5 mol S2O3^2- Then: moles S2O3^2- / liters Na2S2O3 = [Na2S2O3] 9.42 x 10^-5 / [17.64 mL x (1 L / 1000 mL)] = 0.00534 M

Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000157 mol KIO3 were titrated with an unknown solution of Na2S2O3 and the endpoint was reached after 14.63 mL. What is the concentration of the Na2S2O3 solution, in M?

0.00644 Redox reaction: IO3^− (aq) + 6S2O3^2− (aq) + 6H+ (aq) ⟶ I− (aq) + 3S4O6^2− (aq) Equation: (moles of KIO3) x (6 mol S2O3^2- / 1 mol IO3^-) = mol S2O3^2- (0.0000157 mol) x (6/1) = 9.42 x 10^-5 mol S2O3^2- Then: moles S2O3^2- / liters Na2S2O3 = [Na2S2O3] 9.42 x 10^-5 / [14.63 mL / (1 L / 1000 mL)] = 0.00644 M

A 25.0 mL sample of a saturated Ca(OH)2 solution is titrated with 0.028 M HCl, and the equivalence point is reached after 38.1 mL of titrant are dispensed. Based on this data, what is the concentration (M) of Ca(OH)2?

0.021 M 0.028 M x 38.1 mL x (1 L / 1000 mL) = 0.0010668 mol (1/2) x 0.0010668 mol = 0.0005334 mol (0.0005334 mol / 25.0 mL) x (1000 mL / 1 L) = 0.021 M

A 25.0 mL sample of a saturated Ca(OH)2 solution is titrated with 0.026 M HCl, and the equivalence point is reached after 35.9 mL of titrant are dispensed. Based on this data, what is the concentration (M) of the hydroxide ion?

0.037 M 0.026 M x 35.9 mL x (1 L / 1000 mL) = 0.0009334 mol (1/2) x 0.0009334 mol = 0.0004667 mol (0.0004667 mol / 25.0 mL) x (1000 mL / 1 L) = 0.018668 M 0.018668 M x 2 = 0.037 M

How many moles of HCl are in 53.1 mL of a 1.26 M HCl solution?

0.0669 M = moles / V ⟶ Rearrange: moles = M x V 0.0669 = 1.26 M x 53.1 mL x (1 L / 1000 mL)

How many moles of HCl are in 47.3 mL of a 1.65 M HCl solution?

0.078 M = moles / V ⟶ Rearrange: moles = M x V 0.078 = 1.65 M x 47.3 mL x (1 L / 1000 mL)

Suppose you have 450.0 mL of a 0.250 M sodium hydroxide solution. How many moles of sodium hydroxide are in the solution?

0.113 moles of NaOH (450.0 mL x 0.250 M) / 1000 = 0.113 moles

Suppose you have 300.0 mL of a 0.450 M sodium hydroxide solution. How many moles of sodium hydroxide are in the solution?

0.135 moles of NaOH (300.0 mL x 0.450 M) / 1000 = 0.135 moles

Suppose you have 350.0 mL of a 0.450 M sodium hydroxide solution. How many moles of sodium hydroxide are in the solution?

0.158 moles of NaOH (350.0 mL x 0.450 M) / 1000 = 0.158 moles

Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar. Suppose you prepare a glucose solution using the described procedure. Step 1: Dissolve 276.0 g of glucose in enough water to make 500.0 mL of solution. Step 2: Transfer 23.1 mL of the solution to a new flask and add enough water to make 250.0 mL of dilute solution. What is the concentration (in M) of the glucose solution at the end of the procedure?

0.283 M 276.0 g / 180.16 g/mol = 1.532 mol 1.532 mol / 0.5000 L = 3.064 mol (3.064 mol x 23.1 mL) / 1000 = 0.07078 mol (0.07078 mol / 250.0 mL) x 1000 = 0.283 M

Suppose you are titrating vinegar, which is an acetic acid solution, of unknown strength with sodium hydroxide according to the equation: HC2H3O2 + NaOH ⟶ H2O + NaC2H3O2 If you require 32.17 mL of 0.116 M NaOH solution to titrate 10.0 mL of HC2H3O2 solution, what is the concentration of acetic acid in the vinegar? (M)

0.373 M 0.116 M x 32.17 mL = 3.732 mol 3.732 mol / 10.0 mL = 0.373 M

If the freezing point depression for a solution is 2.5°C and Kf = 4.5°C/m, what is the molality of the solution?

0.56 m 2.5°C / (4.5°C/m) = 0.56 m

Consider the reaction corresponding to a voltaic cell and its standard cell potential: Zn (s) + Cu^2+ (aq) ⟶ Cu (s) + Zn^2+ (aq) E°cell = 1.1032 V What is the cell potential for a cell with a 2.468 M solution of Zn^2 (aq) and 0.1306 M solution of Cu^2+ (aq) at 430.9 K?

1.0481 Ecell = E°cell − (RT / nF) lnQ Ecell = 1.1032 V - (8.314 J/mol-K x 430.9 K) / (2 mol e- x 95485 C/mol e-) In(2.468 / 0.1306) = 1.0481

Consider the reaction corresponding to a voltaic cell and its standard cell potential: Zn (s) + Cu^2+ (aq) ⟶ Cu (s) + Zn^2+ (aq) E°cell = 1.1032 V What is the cell potential for a cell with a 2.263 M solution of Zn^2 (aq) and 0.1862 M solution of Cu^2+ (aq) at 405.6 K?

1.0591 Ecell = E°cell − (RT / nF) lnQ Ecell = 1.1032 V - (8.314 J/mol-K x 405.6 K) / (2 mol e- x 95485 C/mol e-) In(2.263 / 0.1862) = 1.0591

A saturated AgCl solution was analyzed and found to contain 1.25 x 10^-5 M of Ag+ ions. Use this value to calculate the Ksp of AgCl.

1.56E-10 AgCl ⇌ Ag^+ + Cl^− Ksp = [Ag+] [Cl−] Ksp = (1.25 x 10^−5 M) (1.25 x 10^−5 M) = 1.56 x 10^−10

What is the standard cell potential (V) for the spontaneous voltaic cell formed from the given half-reactions? Anode: Reduction Half-Reaction = Fe^2+ (aq) + 2e^− ⟶ Fe (s) E°red (V) = -0.41 Cathode: Reduction Half-Reaction = Cl2 (g) + 2e^− ⟶ 2Cl− (aq) E°red (V) = 1.36

1.77 E°cell = E°red at cathode − E°red at anode E°cell = 1.36 V − (−0.41 V) = 1.77

Suppose you need to prepare 120.0 mL of a 0.271 M aqueous solution of NaCl. What mass, in grams, of NaCl do you need to use to make the solution?

1.90 g 0.271 M x 0.1200 L = 0.03252 0.03252 x 58.5 g/mol = 1.90 g

Suppose that you add 21.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a Kf of 5.12 °C/m. With the added solute, you find that there is a freezing point depression of 3.20 °C compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound?

136 g/mol ΔTf = Kf x m ⟶ Reaarange: m = ΔTf / Kf 3.20 °C / 5.12 °C/m = 0.625 m 0.625 m x 0.250 kg = 0.156 mol 21.2 g / 0.156 mol = 136 g/mol

What mass, in grams, of NaCl needs to be added to 1.6 kg of water in order to create a solution with a freezing point of -5.7 °C? The freezing point depression constant of water is 1.86 °C/m.

143 g 0 - (-5.7 °C) = 5.7 °C M2 = (i x Kf x w2 x 1000) / (∆Tf x w2) M2 = molar mass of the solute Kf = Freezing point depression constant w2 = weight of solute w1 = weight of solvent ∆Tf = Depression in freezing point i = van't Hoff factor 58.5 = (2 x 1.86 °C/m x w2 x 1000) / (5.7 °C x 1600 g) 58.5 = (3720 x w2) / (9120) 58.5 = 0.4079 x w2 w2 = 58.5 / 0.4079 w2 = 143 g

What is the heat, q, in joules transferred by a chemical reaction to the reservoir of a calorimeter containing 125 g of dilute aqueous solution (c = 4.184 J/g⋅°C) if the reaction causes the temperature of the reservoir to rise from 21.5°C to 24.5°C?

1569 q = m x c x ΔT q = (125 g) x (4.184 J/g⋅°C) x (24.5°C - 21.5°C) = 1569

Suppose that you add 27.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a Kf of 5.12 °C/m. With the added solute, you find that there is a freezing point depression of 3.56 °C compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound?

159 g/mol ΔTf = Kf x m ⟶ Reaarange: m = ΔTf / Kf 3.56 °C / 5.12 °C/m = 0.695 m 0.695 m x 0.250 kg = 0.174 mol 27.6 g / 0.174 mol = 159 g/mol

What is the correct reading of the volume in the pictured buret? Make sure to report your reading with the appropriate significant figures. (Acid/Base Titrations, Buffers, and pH)

17.32 mL

Calculate the percent by mass of a solution made from 15 g NaCl (the solute) and 67 g water.

18% % mass of solution = (mass of solute/mass of solution) x 100 18% = (15 g / 15 g + 67 g) x 100

Calculate the percent by mass of a solution made from 14 g NaCl (the solute) and 61 g water.

19% % mass of solution = (mass of solute/mass of solution) x 100 19% = (14 g / 14 g + 61 g) x 100

A 250 mL sample of saturated AgOH solution was titrated with HCl, and the endpoint was reached after 2.60 mL of 0.0136 M HCl was dispensed. Based on this titration, what is the Ksp of AgOH?

2.0E-8 0.00260 L x 0.0136 M = 3.54 x 10^−5 mol (3.54 x 10^−5 mol / 0.250 L) = 1.41 x 10^−4 M Ksp = [Ag+] [OH−] Ksp = (1.41 x 10^−4 M) (1.41 x 10^−4 M) = 2.00 x 10^−8

Suppose a student titrated vinegar, which is an acetic acid solution, of unknown strength with a sodium hydroxide solution according to the following equation: HC2H3O2 + NaOH ⟶ H2O + NaC2H3O2 If 14.46 mL of 0.1506 M NaOH were required to titrate 5.00 mL of HC2H3O2. What is the percent (mass/volume) of the vinegar? The molar mass of acetic acid is 60.1 g/mol.

2.62% moles HC2H3O2 = moles NaOH = L NaOH x (moles NaOH / 1 L) [14.46 mL x (1 L / 1000 mL)] x (0.1506 mol / 1 L) = 0.002178 0.002178 mol x 60.1 g/mol = 0.1309 g (0.1309 g / 5.00 mL) x 100 = 2.62%

Suppose a student titrated vinegar, which is an acetic acid solution, of unknown strength with a sodium hydroxide solution according to the following equation: HC2H3O2 + NaOH ⟶ H2O + NaC2H3O2 If 14.67 mL of 0.1506 M NaOH were required to titrate 5.00 mL of HC2H3O2. What is the percent (mass/volume) of the vinegar? The molar mass of acetic acid is 60.1 g/mol.

2.66% moles HC2H3O2 = moles NaOH = L NaOH x (moles NaOH / 1 L) [14.67 mL x (1 L / 1000 mL)] x (0.1506 mol / 1 L) = 0.002209 0.002209 mol x 60.1 g/mol = 0.1328 g (0.1328 g / 5.00 mL) x 100 = 2.66%

If 0.98 g of an unknown was dissolved in 10.30 g of solvent and the resulting solution has a molality of 0.45 m, what is the molar mass of the unknown?

211 g/mol molality = mol solute / kg solvent ⟶ Rearrange: mol solute = molality x kg solvent 0.45 m x 0.0130 kg = 0.004563 mol solute Then: 0.98 g / 0.004563 mol = 211 g/mol

What is the correct reading of the volume in the pictured buret? Make sure to report your reading with the appropriate significant figures. (Determination of the Amount of Sodium Hypochlorite in Commercial Bleach: A Redox Titration)

25.60 mL

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calculate the volume (in mL) of the 1.400 M stock NaOH solution needed to prepare 250.0 mL of 0.1476 M dilute NaOH solution.

26.36 mL NaOH M1 x V1 = M2 x V2 ⟶ Rearrange: V1 = (M2 x V2) / M1 26.36 mL = (0.1476 M x 250.0 mL) / 1.400 M

Suppose you need to prepare 131.1 mL of a 0.473 M aqueous solution of NaCl. What mass, in grams, of NaCl do you need to use to make the solution?

3.62 g 0.473 M x 0.1311 L = 0.062 0.062 x 58.5 g/mol = 3.62 g

What is the correct reading of the volume in the pictured buret? Make sure to report your reading with the appropriate significant figures. (Volumetric Analysis)

39.95 mL

Consider the balanced reversible reaction of acetic acid with ethanol, which takes place with no solvent water: acetic acid + ethanol ⇌ ethyl acetate + water When you react 8.25 M acetic acid with 8.25 M ethanol, the equilibrium concentration of acetic acid is 3.90 M. What is the equilibrium concentration (M) of ethyl acetate?

4.35 [acetic acid]initial − [acetic acid]eq = [ethyl acetate]eq 8.25 M - 3.90 M = 4.35

Consider the balanced reversible reaction of acetic acid with ethanol, which takes place with no solvent water: acetic acid + ethanol ⇌ ethyl acetate + water When you react 8.71 M acetic acid with 8.71 M ethanol, the equilibrium concentration of acetic acid is 3.86 M. What is the equilibrium concentration (M) of ethyl acetate?

4.85 [acetic acid]initial − [acetic acid]eq = [ethyl acetate]eq 8.71 M − 3.86 M = 4.85

Suppose you mix 100.0 g of water at 27.8°C with 75.0 g of water at 76.4°C. What will be the final temperature of the mixed water, in °C?

48.6°C −75.0 g x 4.184 J/g⋅°C x (Tf − 76.4°C) = 100.0 g x 4.184 J/g⋅°C x (Tf − 27.8°C) (plug into calculator; plug in Tf as one letter, e.g. x ) or steps: −75.0 g x 4.184 J/g⋅°C x (Tf − 76.4°C) = 100.0 g x 4.184 J/g⋅°C x (Tf − 27.8°C) 75.0 g x s x (76.4 - T)°C = 100.0 g x s x (T - 27.8)°C 75.0 g x (76.4 - T)°C = 100.0 g x (T - 27.8)°C 5730 - 75.0T = 100.0T - 2780 5730 + 2780 = 100.0T + 75.0T 8510 = 175.0T 8510 / 175.0 = 175.0T / 175.0 48.6°C = T

Suppose you are studying the Ksp of KClO3, which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4.00 g of KClO3 in 12 mL of water at 85°C and cool the solution. At 74°C, a solid begins to appear. What is the Ksp of KClO3 at 74°C?

7.4 (4.00 g / 12 mL) x (1 mol / 122.5 g) x (1000 mL / 1 L) = 2.72 [KClO3] = [K+] = [ClO3^−] = [s] [s] (2.72)^2 = 7.4

Using the procedure in Determination of an Equilibrium Constant, a student combines equimolar amounts of acetic acid and ethanol to react as shown by the following equation: acetic acid + ethanol ⇌ ethyl acetate + water To determine the initial concentration of acetic acid, 1.00 mL of the resulting mixture was titrated with 0.3005 M NaOH, requiring 28.35 mL. What is the molarity of the acetic acid?

8.52 V acid x M acid = V NaOH x M NaOH ⟶ Rearrange: M acid = (V NaOH x M NaOH) / V acid 8.52 = (28.35 mL x 0.3005 M) / 1.00 mL

Suppose you analyze a 30.3 g sample of bleach and determine that there are 2.61 g of sodium hypochlorite present. What is the percent of sodium hypochlorite in the bleach sample?

8.61% % composition = (mass of mixture component / total mass of mixture) x 100 8.61 = (2.61 g / 30.3 g) x 100

Suppose you analyze a 30.2 g sample of bleach and determine that there are 2.66 g of sodium hypochlorite present. What is the percent of sodium hypochlorite in the bleach sample?

8.81% % composition = (mass of mixture component / total mass of mixture) x 100 8.81 = (2.66 g / 30.2 g) x 100

Using the procedure in Determination of an Equilibrium Constant, a student combines equimolar amounts of acetic acid and ethanol to react as shown by the following equation: acetic acid + ethanol ⇌ ethyl acetate + water To determine the initial concentration of acetic acid, 1.00 mL of the resulting mixture was titrated with 0.2950 M NaOH, requiring 31.25 mL. What is the molarity of the acetic acid?

9.22 V acid x M acid = V NaOH x M NaOH ⟶ Rearrange: M acid = (V NaOH x M NaOH) / V acid 9.22 = (31.25 mL x 0.2950 M) / 1.00 mL

In the Determination of an Equilibrium Constant lab, you will react acetic acid with ethanol in the presence of sulfuric acid until equilibrium is established. Then, you will titrate the equilibrium mixture with NaOH to determine the acetic acid concentration. Suppose you determine that the added sulfuric acid requires 2.97 mL of NaOH to be neutralized in a blank solution. The equilibrium mixture requires 12.92 mL of NaOH to reach the endpoint. What volume of NaOH should you use to calculate the acetic acid concentration at equilibrium?

9.95 mL 12.92 mL - 2.97 mL = 9.95 mL of NaOH to react with acetic acid

What mass, in grams, of NaCl needs to be added to 1.5 kg of water in order to create a solution with a freezing point of -4.1 °C? The freezing point depression constant of water is 1.86 °C/m.

97 g 0 - (-4.1 °C) = 4.1 °C M2 = (i x Kf x w2 x 1000) / (∆Tf x w2) M2 = molar mass of the solute Kf = Freezing point depression constant w2 = weight of solute w1 = weight of solvent ∆Tf = Depression in freezing point i = van't Hoff factor 58.5 = (2 x 1.86 °C/m x w2 x 1000) / (4.1 °C x 1500 g) 58.5 = (3720 x w2) / (6150) 58.5 = 0.6049 x w2 w2 = 58.5 / 0.6049 w2 = 97 g

Identify the correct equation for the equilibrium constant K for the reaction given: Cu (s) + 2AgNO3 (aq) ⇌ Cu(NO3)2 (aq) + 2Ag (s)

K = [Cu(NO3)2] / [AgNO3]^2

Identify the correct equation for the equilibrium constant Ka for the weak acid dissociation of acetic acid: HC2H3O2 (aq) + H2O(l) ⇌ H3O+ (aq) + C2H3O2^− (aq)

Ka = [H3O+] [C2H3O2^− ] / [HC2H3O2]

If x represents the molar solubility of Ba3(PO4), what is the correct equation for the Ksp?

Ksp = (3x)^3 (2x)^2 Ba3(PO4)2 ⇌ 3Ba^2+ + 2PO4^3− Ksp = [Ba^2+]^3 [PO4^3−]^2 Ksp = (3x)^3 (2x)^2

If x represents the molar solubility of Ca(OH)2, what is the correct equation for the Ksp?

Ksp = 4x^3 Ksp = [Ca^2+] [OH^−]^2 = (x) (2x)^2 = (x) (4x^2) Ksp = 4x^3

Consider the reversible dissolution of lead(II) chloride: PbCl2 (s) ⇌ Pb^2+ (aq) + 2Cl− (aq) Suppose you dissolve lead(II) chloride in 0.100 M NaCl instead of water. Fill out the ICE table to represent the system if you were asked to calculate amounts at equilibrium. Do not include any units. PbCl2 ⇌ Pb^2+ + 2Cl− IIIIIIIIIIIIIII ___?________?____ CCCCCC ___?________?_____ EEEEEEEE ___?_______?_____

PbCl2 ⇌ Pb^2+ + 2Cl− IIIIIIIIIIIIIII__0.00___0.100__ CCCCCC __+x______+2x__ EEEEEEEE __x______0.100+2x__

Suppose you are studying the kinetics of the reaction between the peroxydisulfate ion and iodide ion. You perform the reaction multiple times with different starting concentrations and measure the initial rate for each, resulting in this table. Experiment 1 : [S2O8^2-] (M) = 0.27 [I^-] (M) = 0.38 Initial Rate (M/s) = 2.05 Experiment 2 : [S2O8^2-] (M) = 0.40 [I^-] (M) = 0.38 Initial Rate (M/s) = 3.06 Experiment 3 : [S2O8^2-] (M) = 0.40 [I^-] (M) = 0.22 Initial Rate (M/s) = 1.76 Based on the data, choose the correct exponents to complete the rate law. Rate = k[S2O8^2-]^a[I^-]^b

a = 1 , b = 1 rate 2 / rate 1 = (concentration 2/ concentration 1)^x 3.06 / 2.05 = (0.40 / 0.27)^a 1.49 = (1.48)^a ; therefore, a = 1 3.06 / 1.76 = (0.38 / 0.22)^b 1.74 = (1.73)^b ; therefore, b = 1

Suppose you are studying the kinetics of the reaction between nitrogen monoxide and hydrogen gas. You perform the reaction multiple times with different starting concentrations and measure the initial rate for each, resulting in this table. Experiment 1 : [NO] (M) = 0.0050 [H2] (M) = 0.0020 Initial Rate (M/s) = 1.25 x 10^-5 Experiment 2 : [NO] (M) = 0.010 [H2] (M) = 0.0020 Initial Rate (M/s) = 5.00 x 10^-5 Experiment 3 : [NO] (M) = 0.0050 [H2] (M) = 0.0040 Initial Rate (M/s) = 2.50 x 10^-5 Based on the data, choose the correct exponents to complete the rate law. Rate = k[NO]^x[H2]^y

x = 2 , y = 1 To study NO, compare experiments 1 and 2. The concentration of H2 stays the same so only NO is changing the rate. Between the two experiments, [NO] is doubled and the rate increase is four-fold. This is a squared impact of concentration, so the rate law must be second order in NO. To study H2, compare experiments 1 and 3. The concentration of NO stays the same so only H2 is changing the rate. Between the two experiments, [H2] is doubled and the rate is also doubled. The rate has the same change as the concentration, so the rate law must be first order in H2.