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3 years ago

14

What is the molarity of a solution made by dissolving 10.305 g of calcium carbonate, CaCO3, in water and diluting with water to

110.0 mL total?

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

0.9359 M

Explanation:

M(CaCO3)=100.1 g/mol

10.305 g * 1 mol/100.1 g = 10.305/100.1 mol CaCO3

(10.305/100.1 mol )/0.1100 L = 0.936 mol/L = 0.9359 M

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Which statement is true about the concentration of the reactants and products in chemical equilibrium?

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Answer:

Ans: 2

Explanation:

The concentration of reactants and the concentration of products are constant.

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What is the molar mass of Ba3(PO4)2?

Yakvenalex [24]

Answer:

<h3>601.93 g/mol</h3>

<h3>explanation:</h3>

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2 years ago

Calculate the frequency of a yellow light with a wavelength of 5.60 x 10-9 m.

Whitepunk [10]

Answer:

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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:

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

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Data:

50/50 ethylene glycol (EG):water

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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

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

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

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

A cylinder of oxygen gas has a pressure of 6.00 atm at 25.0

Readme [11.4K]

P1/T1=P2/T2 Gal Lussac's Law
25 C= 298K (just add 273)
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2 years ago