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3 years ago

What is the scientific tool required to measure the volume of a liquid?

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Volumetric flask? Graduated cylinder? micro-pipets? There's a lot of tools that measure volume

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Nezavi [6.7K]

Try this solution, all the details are described in the attached picture.

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3 years ago

Which statement best describes the development of theories that connected microscopic and macroscopic phenomena?

Alina [70]

The statement which best describes the development of theories that connected microscopic and macroscopic phenomena is; qIt took several hundred years for scientists to develop current theories, and they are still being revised to this.

Discussion;

Most scientific theories involving microscopic and macroscopic phenomenon have taken several years to be developed; however, this theories are still under revision till date.

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2 years ago

Which law was used to determine the relationship between the volume and the number of moles?

lawyer [7]

Answer:

The Mole-Volume Relationship: Avogadro's Law. A plot of the effect of temperature on the volume of a gas at constant pressure shows that the volume of a gas is directly proportional to the number of moles of that gas. This is stated as Avogadro's law.

Explanation:

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2 years ago

Help me I got stuck Pls

Sophie [7]

The answer is C because the weight is going down so they form a gas.

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

7 0

3 years ago