CA salt of strong acid and weak acid
D. A salt of weak acid and weak base
42. What volume of 2.0 M NaOH is required to completely neutralize 25.0 ml, of 2.70 MH.S07
A. 67.5 mL
B. 33.8 mL
C. 22.5 ml.
D. 16.9 mL.
43.
Which of the following statement describes the similarity between voltaic cells and electrolytic cells?
A. Reduction half-reaction occurs at the anode in both types of cells.
B. Both types of cells contain two electrodes in contact with electrolytes
C. The anode is positive and cathode is negative in both types of cells. D. Oxidation half-reaction occurs at the cathode in both types of cella.
44.
WITTE LO
Which of the following occurs during electrolysis of the molten binary salt of a metal? A. Reduction will take place at the anode
B. Oxidation will take place at the cathode
C. No current will flow through the molten salt D. The metal in the salt will deposit on the cathode
45. Which one of the following statements is FALSE regarding the electrolysis of dilute sulfuric acid solution?
A. Oxygen gas is produced at the anode.
B. Hydrogen gas is formed at the cathode
C. At the anode, SOions are discharged
D. At the cathode, H ions are discharged
46. Which of the following statement is FALSE regarding electrolysis?
A. Inert electrode is an electrode that does not affect the product of electrolysis.
B. Active or reactive electrode is an electrode that affects the product of electrolysis.
C. Preferential discharge is the ease in which ions are discharged in preference of other ions during D. The electrical conduction of electrolytes is due to the movement of delocalized electrons.
electrolysis.
47. Calculate the mass of copper deposited by the electrolysis of copper (II) sulfate solution placed in series with a silver nitrate solution, when 0.108 g of silver is being deposited. (Molar mass of Ag=108 g/mol and molar
mass of Cu-63.5 g/mol)
A. 0.066 g
C. 0.3178
D. 0.635 g
B. 0.032 g 48. What mass of chlorine gas can be produced when a curent of 0.452 A is passed through molten CaCh fes
1.5 hours? (IF-96,500 Cmol)
A. 1018
B. 0.507
C. 0.898z
D 1.795
A. 0.066 g
48. A. 1018
42. To solve this problem, we can use the concept of stoichiometry and the equation for the reaction between NaOH and H2SO4: 2 NaOH + H2SO4 → Na2SO4 + 2 H2O
First, we need to find the number of moles of H2SO4 using the given concentration and volume:
moles H2SO4 = molarity × volume
moles H2SO4 = 2.70 M × 25.0 mL
Next, we can use the stoichiometry of the balanced equation to find the number of moles of NaOH required to neutralize the H2SO4. From the equation, we can see that the ratio of moles of H2SO4 to NaOH is 1:2.
moles NaOH = 2 moles NaOH/1 mole H2SO4 × moles H2SO4
Finally, we can use the new moles of NaOH to find the volume of 2.0 M NaOH required to neutralize the H2SO4 using the equation:
volume NaOH = moles NaOH / molarity NaOH
Calculate:
moles H2SO4 = 2.70 M × 25.0 mL = 0.0675 moles H2SO4
moles NaOH = 2 moles NaOH/1 mole H2SO4 × 0.0675 moles H2SO4 = 0.135 moles NaOH
volume NaOH = 0.135 moles NaOH / 2.0 M = 0.0675 L = 67.5 mL
Therefore, the correct answer is A. 67.5 mL.
43. The correct answer is B. Both types of cells contain two electrodes in contact with electrolytes.
44. The correct answer is D. The metal in the salt will deposit on the cathode.
45. The correct answer is C. At the anode, SO4 ions are discharged.
46. The correct answer is D. The electrical conduction of electrolytes is due to the movement of delocalized electrons.
47. To solve this problem, we can use Faraday's laws of electrolysis and the equation Q = nF, where Q is the charge passed, n is the number of moles, and F is the Faraday constant (96500 C/mol). We can set up a proportion using the ratio of the charges of Ag and Cu:
charge Cu / charge Ag = moles Cu / moles Ag
Substituting the given values into the equation:
0.108 g Ag × (1 mol Ag / 108 g Ag) × (2 mol Cu / 1 mol Ag) = mass Cu
Calculate:
mass Cu = 0.108 g Ag × (1 mol Ag / 108 g Ag) × (2 mol Cu / 1 mol Ag) = 0.001 g Cu = 0.001 g
Therefore, the correct answer is B. 0.001 g.
48. To solve this problem, we can use Faraday's laws of electrolysis and the equation Q = nF, where Q is the charge passed, n is the number of moles, and F is the Faraday constant (96500 C/mol). First, we need to calculate the charge passed using the equation:
charge = current × time
Calculate:
charge = 0.452 A × 1.5 hours × 3600 s/h = 2446.4 C
Next, we can use the charge passed to calculate the number of moles of Cl2 using Faraday's laws:
n = charge / F
Calculate:
n = 2446.4 C / 96500 C/mol = 0.0254 mol Cl2
Finally, we can convert moles of Cl2 to mass using the molar mass of chlorine gas:
mass = moles × molar mass
Calculate:
mass = 0.0254 mol Cl2 × 70.9 g/mol = 1.798 g
Therefore, the correct answer is D. 1.798 g.