Which of the following items represents a chemical change?

A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

5 0

3 years ago

Write the correct answers.

Ilya [14]

Answer:

pour in hot water and It will disolve

7 0

3 years ago

4NH3 + 502 --> 4NO + 6H20

gregori [183]

Answer:

5.52 g

Explanation:

4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert the given masses of both reactants into moles, using their respective molar masses:

6.30 g NH₃ ÷ 17 g/mol = 0.370 mol NH₃

1.80 g O₂ ÷ 32 g/mol = 0.056 mol O₂

Now we calculate with how many NH₃ moles would 0.056 O₂ moles react, using the stoichiometric coefficients:

0.056 mol O₂ * $\frac{4molNH_3}{5molO_2}$ = 0.045 mol NH₃

As there more NH₃ moles than required, NH₃ is the excess reactant.

Then we calculate how many NH₃ moles remained without reacting:

0.370 mol NH₃ - 0.045 mol NH₃ = 0.325 mol NH₃

Finally we convert NH₃ moles into grams:

0.325 mol NH₃ * 17 g/mol = 5.52 g

7 0

3 years ago

What is a heterogenous mixture?

Nezavi [6.7K]

Answer: a heterogenous mixture is simply any mixture that is not uniform in composition- its non-uniform mixture of smaller constituent parts

Explanation:

8 0

3 years ago

An insoluble solid that forms from a chemical reaction is called a(n)

Kisachek [45]

This description applies and is suitable for what a chemical precipitate is. A precipitate is a product that is formed from a certain chemicals reaction that yields a solid that is insoluble in the reaction vessel. It is usually white and opaque.

4 0

3 years ago