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KIM [24]
3 years ago
9

n atom “likes” to obtain a full outer shell. The fourth shell (in the atoms that we are studying today) can hold a maximum of 8

electrons. Think: Energetically, would the atom above “prefer” to give away its valence electrons or borrow some from another atom in order to fill its outer shell? Explain your reasoning
Chemistry
1 answer:
fgiga [73]3 years ago
4 0

It depends on how many electrons are already in the outer shell.

If it is a large amount (if it is almost to the maximum), then it will want to borrow from another atom.

But if it is close to empty then it will want to give away so it will go back to the inner ring with will be full.

Hope this helps, good luck!

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what mass of lithium chloride, licl must be dissolved to make a 0.194M solution that has volume of 1.00 l
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MXV= (0.194M)(1.00L)=0.194moles
42.39LiClg/molex0.194moles=8.2g LiCl
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When boiling water is the particles expanding?
Juli2301 [7.4K]
When water is boiling the particles in the liquid are not expanding. This also applies for other liquids. Particles do not expand, it is only the volume they take up that expands. When water is boiling, the particles are rather escaping the liquid phase by undergoing a phase change which forms the particles into the gaseous phase.
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3 years ago
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.

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Potential energy or kinetic energy.
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On a very hot day, could you cool your kitchen by leaving the refrigerator door open?
Scilla [17]

Answer:c

Explanation:

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