Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
<span>1s2, 2s2, 2p6, 3s2, 3p6, 3d5</span>
Answer: 0.0250
Explanation: 10 X 0.0750 = .75
.75 / 30 = 0.0250 M
Put the salt and sand in some water.
the salt will not be visible but the sand will
now strain the the sand from the water
now boil the water and now the water will boil away and now you will just have salt left.
