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lbvjy [14]
3 years ago
6

How many mercury atoms are there in 3.8 mol of mercury? Show work please

Chemistry
1 answer:
netineya [11]3 years ago
4 0
There are 6.02x10^22 molecules of any atom/molecule in one mole of anything.

Therefore no. of atoms in 3.8 mols of Mercury= 3.8 * 6.02x10^22=  2.2876x10^23 atoms
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Q2. Name, constituent of petroleum (Petrochemicals) which is used for following purposes.
finlep [7]

The constituents of petroleum that are used for the following purposes are as follows:

  • To make candles ----- Paraffin wax
  • A solvent for dry cleaning ----- Petrol
  • For surfacing roads ----- Bitumen
  • Jet engine fuel ----- Kerosene
  • For lubrication ----- Lubricating oil.

<h3>What are the constituents of petroleum?</h3>

The constituents of petroleum are LPG, bitumen, paraffin wax, lubricating oil, kerosene, diesel, etc. These compounds are a mixture of hydrocarbons.

Therefore, each constituent of petroleum that is used for the following purposes is mentioned above with proper names.

To learn more about constituents of petroleum, refer to the link:

brainly.com/question/17023299

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7 0
1 year ago
What type of compound is always an electrolyte?
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All ionic compounds
<span>Ionic compounds always act as an electrolyte.As soon as they are added to water to form solutions, they dissociate in their constituent ions.</span>
6 0
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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How are the structure and properties of solids related
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The general properties of solids reflect the orderly arrangement of their particles and the fixed locations of their particles.

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What are two things that characterize the practice of science
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1. Curiosity
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