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Triss [41]
3 years ago
11

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

ng system holds 4.70 gal, what is the boiling point of the solution? For the calculation, assume that at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Also, assume that the engine coolant is pure ethylene glycol ( HOCH 2 CH 2 OH ) , which is non‑ionizing and non‑volatile, and that the pressure remains constant at 1.00 atm. The boiling‑point elevation constant for water will also be needed.
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

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. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid =  diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol

5 0
3 years ago
Which of the following trends can be observed when moving down a column in the periodic table
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Explanation:

A column on the periodic table represents a period.

These are some of the variations observed across a period;

  • Atomic radii decreases progressively from left to right due to the progressive increase in the nuclear charge without an attendant increase in the number of electronic shells.
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A volume of 10.0L of gas at a temperature of 5c is cooled to a temperature of 85C at constant pressure what is the new volume of
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I don't know how 5°C cooled to 85°C but the answer would be 12.878L

4 0
3 years ago
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The balanced reaction between aqueous nitric acid and aqueous strontium hydroxide is _________
umka2103 [35]

Answer:

2HNO_3(aq)+Sr(OH)_2(aq)\rightarrow Sr(NO_3)_2(aq)+2H_2O(l)

Explanation:

Hello,

In this case, since nitric acid is HNO₃ and strontium hydroxide is Sr(OH)₂ we can represent the balanced chemical reaction by equaling the atoms of strontium, nitrogen, oxygen and hydrogen at both reactants and products as shown below:

2HNO_3(aq)+Sr(OH)_2(aq)\rightarrow Sr(NO_3)_2(aq)+2H_2O(l)

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