Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Explanation:
I THINK
They do form directly on the equators
Answer:
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Explanation:

Moles of copper = 
According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.
Then 0.03613 moles of copper will give:
of nitrogen dioxide gas
Moles of nitrogen dioxide gas = n = 0.06326 mol
Pressure of the gas = P
P = Total pressure - vapor pressure of water
P = 726 mmHg - 23.8 mmHg = 702.2 mmHg
P = 0.924 atm (1 atm = 760 mmHg)
Temperature of the gas = T = 25.0°C =298.15 K
Volume of the gas = V


V = 1.68 L
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
A water molecule, because of its shape, is a polar molecule.