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3 years ago

Use the table to calculate ΔH°rnx for the following reaction

Chemistry

1 answer:

alexdok [17]3 years ago

7 0

Answer:

The answer to your question is ΔH° rxn = -1343.9 kJ/mol

Explanation:

P₄O₆ (s) + 2 O₂ (g) ⇒ P₄O₁₀

ΔH°rxn = ?

Formula

ΔH°rxn = ∑H° products - ∑H° reactants

H° P₄O₆ = -1640.1 kJ/mol

H° O₂ = 0 kJ/mol

H° P₄O₁₀ = -2984 kJ/mol

-Substitution

ΔH° rxn = (-2984) - (-1640.1) - (0)

-Simplification

ΔH° rxn = -2984 + 1640.1

ΔH° rxn = -1343.9 kJ/mol

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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

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