I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
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
Your answer is radio waves
Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
