Answer:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Explanation:
Step 1: The balanced equation
2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)
This equation is balanced, we do not have the change any coefficients.
Step 2: The netionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)
After canceling those spectator ions in both side, look like this:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Answer:
C.) 2
Explanation:
The pH equation is:
pH = -log[H⁺]
In this equation, [H⁺] is the molarity of the acid. In this case, the acid is HCl. Molarity can be found using the equation:
Molarity (M) = moles / volume (L)
Since you were given moles and volume, you can find the molarity of HCl.
Molarity = moles / volume
Molarity = 0.01 moles / 1.00 L
Molarity = 0.01 M
Now, you can plug the molarity of the acid into the pH equation.
pH = -log[H⁺]
pH = -log[0.01]
pH = 2
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
It will float, because it is almost weighs like a pen.
Cs+1
The only common oxidation state is +1.
