Need help asap please and thank you!!

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

7 0

3 years ago

I Need help asap i will give brainliest

Flauer [41]

Answer:

It maintains a constant internal temperature.

Explanation:

Our body tries its hardest to maintain a constant internal temperature. This is because if we get to warm or cold it is unhealthy. If your body senses that it is getting to warm or cold it will try to correct itself.

3 0

3 years ago

Which of the following best helps to explain why the electron affinity of Br has a greater magnitude than that of I?

tankabanditka [31]

The reason why Br has a greater magnitude of electron affinity than that of I is that there is a greater attraction between an added electron and the nucleus in Br than in I.

In the periodic table, there are trends that increase down the group and across the period. Electron affinity is a trend that increases across the period but decreases down the group.

Recall that the ability of an atom to accept an electron depends on the size of the atom. The smaller the atom, the greater the attraction between an added electron and the nucleus.

Since Br is smaller than I, there is a greater attraction between an added electron and the nucleus in Br than in I which explains why Br has a greater magnitude of electron affinity than I.

Learn more: brainly.com/question/17696329

7 0

3 years ago

Which example describes how the water cycle carries on energy transfer? a flooding river depositing silt on a floodplain a warm

Irina18 [472]

Answer:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

5 0

3 years ago

Read 2 more answers

Hydrogen can be extracted from natural gas according to the following equilibrium.

antiseptic1488 [7]

Answer:

2.16x10⁻²

Explanation:

First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:

nCH4 = 24.0/16 = 1.5 moles

nCO2 = 88.0/44 = 2 moles

The concentration is the number of moles divded by the volume, thus:

[CH4] = 1.5/1 = 1.50 M

[CO2] = 2/1 = 2.00 M

For the equilibrium reaction, let's do an equilibrium chart:

CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)

1.50 2.00 0 0 Initial

-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)

1.50-x 2.00-x 2x 2x Equilibrium

As sateted in problem, [CH4] = 2.70*[CO]

1.50 - x = 2.70*2x

1.50 - x = 5.4x

6.4x = 1.50

x = 0.2344

Thus, at equilibrium:

[CH4] = 1.50 - 0.2344 = 1.2656 M

[CO2] = 2.00 - 0.2344 = 1.7656 M

[CO] = 2*0.2344 = 0.4688 M

[H2] = 2*0.2344 = 0.4688 M

The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.

K = ([CO]²*[H2]²)/([CH4]*[CO2])

K = (0.4688²*0.4688²)/(1.2656*1.7656)

K = 0.0483/2.2345

K = 2.16x10⁻²

7 0

3 years ago