The formula of the hydrate = CuSO₄• 3H₂O
<h3>Further explanation</h3>
Given
4.175 grams sample CuSO₄• xH₂O
3.120 grams anhydrous compound CuSO₄
Required
The formula
Solution
mass of H₂O driven off :
= 4.175 - 3.12
= 1.055 g
MW CuSO₄ = 159.5 g/mol
MW H₂O = 18 g/mol
mol ratio of CuSO₄ : H₂O :
= 3.12/159.5 : 1.055/18
= 0.01956 : 0.05861
= 1 : 3
the solid particles take up the intermolecular spaces in the liquid.
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
Answer: when concentrations of acid and base are same, pH = pKa
PH = 12.38 pOH = 1.62
Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00
