Density is the ratio of a substance ___ to its volume A. Force B. Mass C. Thickness D. Weight

Chemistry

1 answer:

Svetach [21]3 years ago

3 0

Answer:

Density is the ratio of a substance mass to its volume

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How many grams of CaCl2 are needed to make 277.8g of a solution that is 31.5% (m/m) in water? Note that mass is not technically

Sveta_85 [38]

Answer: The mass of calcium chloride present in given amount of solution is 87.5 g

Explanation:

We are given:

Mass of solution = 277.8 grams

Also, 31.5 % (m/m) of calcium chloride in water. This means that 31.5 g of calcium chloride is present in 100 g of solution.

To calculate the mass of calcium chloride in the given amount of solution, we use unitary method:

in 100 g of solution, the mass of calcium chloride present is 31.5 g

So, 277.8 g of solution, the mass of calcium chloride present is $\frac{31.5}{100}\times 277.8=87.5g$

Hence, the mass of calcium chloride present in given amount of solution is 87.5 g

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3 years ago

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 $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$