Answer:
ΔHr = -103,4 kcal/mol
Explanation:
<u>Using:</u>
<u>AH° (kcal/mol)
</u>
<u>Metano (CH)
</u>
<u>-17,9
</u>
<u>Cloro (CI)
</u>
<u>tetraclorometano (CCI)
</u>
<u>- 33,3
</u>
<u>Acido cloridrico (HCI)
</u>
<u>-22</u>
It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:
ΔHr = (ΔH products - ΔH reactants)
For the reaction:
CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)
The balanced reaction is:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
The ΔH's of formation for these compounds are:
ΔH CH₄(g): -17,9 kcal/mol
ΔH Cl₂(g): 0 kcal/mol
ΔH CCl₄(g): -33,3 kcal/mol
ΔH HCl(g): -22 kcal/mol
The ΔHr is:
-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)
<em>ΔHr = -103,4 kcal/mol</em>
<em></em>
I hope it helps!
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M
Independent Variable
i hope I've helped!
the answer is "There is an outside energy source." in voltaic cells, there is not need of an external source of energy because they involve spontaneous reactions.
in both types of cells, the electrons move from anode to cathode, the anode is where oxidation takes place and cathode for reduction (an ox, red cat). also both have two half reactions.
