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Zolol [24]
3 years ago
14

In the galvanic cell Al(s) ǀ Al3+(aq, 1 M) ǀǀ Cu2+(aq, 1 M) ǀ Cu(s) which of the following changes will increase the cell potent

ial? I. Dilution of the Al3+ solution to 0.001 M II. Dilution of the Cu2+ solution to 0.001 M III. Increasing the surface area of the Al(s) electrode (A) I only (B) II only (C) III only (D) I and III only
Chemistry
2 answers:
cestrela7 [59]3 years ago
4 0

Answer:

(B) II only

Explanation:

dangina [55]3 years ago
3 0
The cathode electrode is copper. This means that the concentration of copper must be increased to increase the cell potential. This can be done by diluting the Al3+ solution. Increasing the surface area will not change the current but will increase the voltage. The answer is (D) I and III only.
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At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene
valina [46]

Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_1=x_1p_1^0 and p_2=x_2P_2^0

where, x = mole fraction in solution

p^0 = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_1+p_2\\p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0

x_{benzene}=x,

x_{toluene}=1-x_{benzene}=1-x,

p_{benzene}^0=75torr

p_{toluene}^0=22torr

p_{total}=40torr

40=x\times 75+(1-x)\times 22

x=0.34

Thus (1-x0 = (1-0.34)=0.66

Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66

8 0
3 years ago
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