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3 years ago

Which of the following statements about metal elements is correct?

Chemistry

1 answer:

Sati [7]3 years ago

4 0

Answer: C. Metals tend to easily lose their valence electrons.

Explanation:

Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.

Example: Gold, potassium etc

$M\rightarrow M^++e^-$

Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.

Example: Sulphur, Chlorine

$N+e^-\rightarrow N^-$

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Answer:0.90889467046888

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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

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

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

(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.

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4 years ago

Given two temperatures, 36 °F and 72 °F, which of the following correctly compares the number of opportunities for particles to

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The correct answer is "Greater at 72 °F " hope it helps

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Answer:

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