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Illusion [34]
3 years ago
12

Please help as soon as possible

Chemistry
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

A. 2.4 x 10¹⁴ J

Explanation:

m (mass) = 2.7 x 10⁻³ kg

c = 3.0 x 10⁸ m/s

Plug these numbers into the equation E=mc²

E=mc²

E= (2.7 x 10⁻³)(3.0 x 10⁸)²

E= 2.43 x 10¹⁴

So, the answer is A. 2.4 x 10¹⁴ J.

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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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The formula for the boiling point elevation ΔTb is

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i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

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\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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