The answer for the question is c.
Answer : The
for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : 
Reduction : 
Now we have to calculate the Gibbs free energy.
Formula used :

where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.

Therefore, the
for this reaction is, -88780 J/mole.
Answer: 40 grams
Explanation:
The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = 93.4J
M = ?
C = 0.129 J/g.C
Φ = 40.4°C - 22.3°C = 18.1°C
Then, Q = MCΦ
Make Mass, M the subject formula
M = Q/CΦ
M = (93.4J) / (0.129 J/g.C x 18.1°C)
M = 93.4J / 2.33J/g
M = 40 g
Thus, the mass of the lead is 40 grams
Answer:
D = 5.3 g/mL
Explanation:
Density = Mass over Volume
D = m/V
Step 1: Define
D = unknown
m = 16 g
v = 3.0 mL
Step 2: Substitute and Evaluate
D = 16 g / 3.0 mL
D = 5.333333333 g/mL
Step 3: Simplify
We have 2 sig figs.
5.333333333 g/mL ≈ 5.3 g/mL