Answer:
2NaClO₄ (aq) + Ni(NO₃)₂ (aq) → 2NaNO₃ (aq) + Ni(ClO₄)₂ (aq)
No reaction
Explanation:
NaClO₄ → Sodium perchlorate
Ni(NO₃)₂ → Nickel (II) nitrate
All the salts from nitrate are soluble salts
All the salts from perchlorate are soluble salts except for the KClO₄
Explanation:
1)  + 7 H_2(g)](https://tex.z-dn.net/?f=%202%20Al%28s%29%20%2B%202%20NaOH%28aq%29%20%2B%206%20H_2O%28l%29%20%5Clongleftrightarrow%202%20Na%5BAl%28OH%29_4%5D%28aq%29%20%2B%207%20H_2%28g%29)
![Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNa%5BAl%28OH%29_4%5D%5D%5E2%2A%5BH_2%5D%5E7%7D%7B%5BNaOH%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
2) 
![Kc=\frac{[H_2SO_4]}{[SO_3]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2SO_4%5D%7D%7B%5BSO_3%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
3) 
![Kc=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
The concentration of Ca2+ ions is half that of the Cl- ions.
The answer is coastal upwelling.
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!
