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2 years ago

What happens to the energy absorbed during an endothermic reaction

Chemistry

1 answer:

LuckyWell [14K]2 years ago

4 0

Answer:

more energy is absorbed when the bonds in the reactants are broken

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A 508-g sample of sodium bicarbonate (NaHCO3) contains how many moles of sodium bicarbonate (NaHCO3)?

Blizzard [7]

molar mass = (22.99) + (1.01) + (12.01) + 3(16.00)

molar mass = 84.01 g/mol

(508g)(1 mol/84.01 g) = 6.0

There are 6.0 moles of sodium bicarbonate

3 0

3 years ago

Donna and Ellis have been married for over 30 years which of the following is likely to be most important in their relationship

OleMash [197]

30 because it is the highest number of marriage

8 0

3 years ago

Give an example of each type of intermolecular forces (dispersion,

Andrei [34K]

I’m not too sure I hope someone answers for you

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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:

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

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

How many molecules of zinc oxide are there in a 2 kg sample?

Novosadov [1.4K]

There are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample. Details about number of molecules can be found below.

<h3>How to calculate number of molecules?</h3>

The number of molecules of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

According to this question, there are 2000g of ZnO in a sample. Zinc oxide has a molar mass of 81.38 g/mol.

no of moles = 2000g ÷ 81.38g/mol

no of moles = 24.57mol

number of molecules = 24.57 × 6.02 × 10²³

number of molecules = 147.95 × 10²³

Therefore, there are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample.

Learn more about number of molecules at: brainly.com/question/11815186

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2 years ago