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2 years ago

Which one of the following produces hydrogen ions in a water solution?

Chemistry

1 answer:

valina [46]2 years ago

8 0

Answer:

A. Acids

Explanation:

An acid is an ionic compound that produces positive hydrogen ions when dissolved in water. Acids taste sour and turn blue litmus paper red. A base is an ionic compound that produces negative hydroxide ions when dissolved in water.

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What is plasma in chemistry

faltersainse [42]

Explanation:

Plasma is an ionized gas, a distinct fourth state of matter. “Ionized” means that at least one electron is not bound to an atom or molecule, converting the atoms or molecules into positively charged ions.but plasma is present in human body too in a liquid form.

hope this helps you

have a nice day

8 0

2 years ago

Which of the following transports information from the eye to the brain?

Marina CMI [18]

Answer:

Explanation:

c is the correct answer

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2 years ago

What is the potential energy of the bowling ball as it sits on top of the building?

Tju [1.3M]

Answer:

Potential energy is energy due to an object's height above the ground.

Potential energy = mass x gravity x height

Kinetic energy is energy due to the motion of the object.

Kinetic energy = 1/2 x mass x velocity²

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3 years ago

Im the only metal who is a liquid at room temperature

mylen [45]

The only metal that is a liquid at room temperature is mercury!!

5 0

3 years ago

Read 2 more answers

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