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3 years ago

What is the formula of the ionic compound expected to form between the elements chlorine and aluminum?

Chemistry

1 answer:

galben [10]3 years ago

3 0

Answer: AlCl3

Explanation:

Aluminum is in group 13 and has a valency of 3. Chlorine is in group 17 and has a valency of 1. When a compound is formed, it reflects the valencies of each of the combining elements. This can be simplified by saying that the combining elements exchange their valencies. Cl gets 3 and All gets 1 forming a compound with the formula shown above.

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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

Which of the following is an example of a behavioral adaptation

Goryan [66]

Migration is an example of behavioral adaptation

8 0

3 years ago

Read 2 more answers

The density of the hydrocarbon in part (a) is 2.0 g l-1 at 50°c and 0.948atm. (i) calculate the molar mass of the hydrocarbon. (

Vedmedyk [2.9K]

1. Answer;

=56 g/mol

Explanation and solution;

PV = nRT

nRT= mass/molar mass (RT)

molar mass = (mass/V ) × (RT/P)

= Density × (RT/P)

Molar mass = 2.0 g/L × (0.0821 × 323 K)/0.948 atm

Molar mass = 56 g/mol

2. Answer;

Molecular mass is C4H8

Explanation;

Empirical mass × n = molar mass

Empirical mass for CH2 = 14 g/mol

Therefore;

56 g/mol = 14 g/mol × n

n = 4

The molecular formula= 4(CH2)

= C4H8

7 0

3 years ago

Scientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below. A pur

Julli [10]

Answer:

dude that doesnt maek sense

Explanation:

6 0

3 years ago

I need help so please help me

Kay [80]

Answer:b

Explanation:

I honestly don’t know if this I right but that would be my guess

3 0

3 years ago