Answer:
NaCl.
Explanation:
In the solution, ZnSe ionizes to 
and 
. Following reaction represents the ionization of ZnSe in solution -
⇄ 
As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.
If we add NaCl to this solution, then we have 
and 
in the solution which will be formed by the ionization of NaCl.
Now, in the solution will react with two ions to form 
as follows -
⇄
Due to this reaction the concentration of will decrease in the solution and more ZnSe can be soluble in the solution.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
Answer:
4.62
so 5
the ratio is 2 na chlorates for 3 O2 so multiply 7 by 2/3
Explanation:
