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aleksklad [387]

3 years ago

9

The Brownian Motion prevents colloid particles from settling out of a solution. Please select the best answer from the choices p

rovided T F

1 answer:

KonstantinChe [14]3 years ago

7 0

The correct answer would be true

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What would be the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg?

Anna [14]

Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)

V = Volume of gas = 15 L

n = number of moles = 0.6

R = gas constant = $0.0821Latm/Kmol$

T =temperature = ?

$T=\frac{PV}{nR}$

$T=\frac{3.02atm\times 15L}{0.0821Latm/K mol\times 0.6mol}=920K$

Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K

8 0

3 years ago

Which substances in the diagram are most unstable? positive values of enthalpy negative values of enthalpy negative values -50 t

NemiM [27]

Answer:

no s e

Explanation:

tampocose

8 0

2 years ago

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

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3 years ago

cricket20 [7]

Answer:

I THINK expose all six plants to green light

Explanation:

8 0

3 years ago

Which statement is a correct expression of the Law of Conservation of Mass?

Phantasy [73]

<span><span>Law of Conservation of Mass - "The total mass after a chemical reaction is exactly the same as the mass before"</span></span><span>
</span>

6 0

2 years ago