Answer:
True
Explanation:
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The greatest molar amount of metal product is obtained from silver (1) nitrate.
Let us determine the mass of metal obtained in each case.
For La^3+;
La^3+ + 3e ----> La
1 mole of La is deposited by 3F
x moles of La is deposited by 10 F
x = 1 × 10/3
x = 3.33 moles
For Zn^2+;
1 mole Zn^2+ is deposited by 2F
x moles of Zn^2+ is deposited by 10 F
x = 5 F
For Ag^+
1 mole of Ag^+ is deposited by 1 F
x moles of Ag^+ is deposited by 10 F
x = 10 moles
For Ba^2+;
1 mole of Ba^2+ is deposited by 2 F
x moles of Ba^2+ is deposited by 10 F
x = 5 moles
Learn more: brainly.com/question/967776
Answer:
2.0 V
Explanation:
For the oxidation half cell;
Al(s) -------> Al^3+(aq) + 3e.
For reduction half cell;
Cu^2+(aq) +2e ------> Cu(s).
E°cell = E°cathode - E°anode
But;
E°cathode= 0.34 V
E°anode = -1.66 V
E°cell= 0.34 -(-1.66)
E°cell= 2.0 V
Answer : The work done on the surroundings is, 709.1 Joules.
Explanation :
The formula used for isothermally irreversible expansion is :

where,
w = work done
= external pressure = 1.00 atm
= initial volume of gas = 1.00 L
= final volume of gas = 8.00 L
Now put all the given values in the above formula, we get :



The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.
Therefore, the work done on the surroundings is, 709.1 Joules.