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mario62 [17]
3 years ago
11

Calculate the energy needed to raise the temperature of 165.0 g of water from 10.0°C to 40.0°C. The specific heat capacity for w

ater is 4.2J/g °C.
Chemistry
1 answer:
mr Goodwill [35]3 years ago
7 0

Answer:

20790 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 165 g.

Initial temperature (T1) 10 °C.

Final temperature (T2) = 40 °C.

Specific heat capacity (C) = 4.2 J/g °C.

Heat (Q) required =?

Next, we shall determine the change in temperature of water. This can be obtained as illustrated below:

Initial temperature (T1) 10 °C.

Final temperature (T2) = 40 °C.

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 40 – 10

ΔT = 30 °C

Finally, we shall determine the heat energy required to raise the temperature of the water as follow:

Mass (M) = 165 g.

Specific heat capacity (C) = 4.2 J/g °C.

Change in temperature (ΔT) = 30 °C

Heat (Q) required =?

Q = MCΔT

Q = 165 × 4.2 × 30

Q = 20790 J

Thus, the heat energy required to raise the temperature of the water is 20790 J

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(b) What is the pH of 0.40 M triethylammonium chloride, CH₃ (CH₂)₃ NHCl ?
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pH of 0.40M triethylammonium chloride is 5.90.

<h3>What is pH?</h3>

A solution's acidity may be determined by looking at its pH, which is a measurement of hydrogen ion concentration. Pure water slightly separates into ions with roughly equal amounts of hydrogen and hydroxyl (OH) ions. [H+] is 107 for a neutral solution, or pH = 7.

<h3>Given : </h3>

Concentration of triethylammonium chloride = 0.40M

pH = ?

<h3>Solution: </h3>

(CH3CH2)3NHCl ------> (CH3CH2)3NH⁺ + Cl⁻

(CH3CH2)3NH⁺ will react with water to give H3O⁺ .

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1 year ago
IF 3.25 mol of argon gas occupies a volume of 100. L at a particular temperature and
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Answer:

435.38 L

Explanation:

From the question given above, the following data were obtained:

Initial mole (n₁) = 3.25 mole

Initial volume (V₁) = 100 L

Final mole (n₂) = 14.15 mole

Final volume (V₂) =?

The final volume occupied by the gas can be obtained as follow:

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100 / 3.25 = V₂ / 14.15

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V₂ = 435.38 L

Thus, the final volume of the gas is 435.38 L

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