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Marta_Voda [28]
3 years ago
9

How many moles in 68.1 grams of Cu(OH)2

Chemistry
1 answer:
Andrew [12]3 years ago
6 0

To do this, you would first add together the molar mass of all involved elements, to find how many grams are in a mole of Cu(OH)2. Keep in mind, the molar mass is equal to the atomic mass of an element in grams. For example the molar mass of copper (Cu) would be 63.55 (with 2 sig. figs.)

Therefore, now we add together the mass of all elements involved.

Cu: (63.55)+O2(15.99x2=31.98)+H2(1.01x2=2.02)

63.55+31.98+2.02= 97.55g per mole of Cu(OH)2.

Now, divide what we have by how much it takes to get a mole of the stuff.

68.1/97.55= 0.698mol Cu(OH)2


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4.65 g x 1 mol/4.0026 g/mol = 1.162 mol helium

Explanation:

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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
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Answer:

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

Explanation:

Data:

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V = 4.70 gal

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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}

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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Explanation

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