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Anit [1.1K]
3 years ago
14

8. The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to:

Chemistry
1 answer:
dedylja [7]3 years ago
8 0

The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to "add water, stir, filter and evaporate the filtrate".

<u>Option: C</u>

<u>Explanation:</u>

The salt and sand can be separated on the basis of solubility, as we know the salt is chemically known as sodium chloride, which have good solubility in water. The most general method is the mixture is taken into a flask or beaker and water is added with stirring, where salt get dissolved and sand remain as it is, because NaCl is capable enough to form hydrogen bonding with water, while sand have absence of such property. Then this solution containing insoluble sand is filtered by using filter paper. The sand is received in filter paper while filtrate in beaker is evaporated by boiling it in order to receive salt as residue.

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A compound is formed when 9.03 g
NikAS [45]

The percent composition  of this compound :

Mg = 72.182%

N = 27.818%

<h3>Further explanation</h3>

Given

9.03 g  Mg

3.48 g  N

Required

The percent composition

Solution

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

Total mass of the compound :

= 9.03 g + 3.48 g

= 12.51 g

The percent composition :

Mg : 9.03/ 12.51 g x 100% = 72.182%

N : 3.48 / 12.51 g x 100% = 27.818%

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Read 2 more answers
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
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