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solong [7]
3 years ago
10

14. All the following are necessary parts of a neutralization reaction except A. an indicator. B. a salt. C. water. D. an acid.

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
6 0

The correct answer is <em>B. a Salt </em><em>because The reaction of an acid and a base is called a neutralization reaction because the properties of both the acid and base are diminished or neutralized when they react. A neutralization reaction is a reaction of an acid with a base in aqueous solution to produce water and a salt, as shown by the following equation:</em>

<em>acid + base → salt + water</em>

<em />

<em>* Hopefully this helps:) Mark me the brainliest:) </em>

<em>∞ 234483279c20∞</em>

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Answer:

Explanation:

Well, obviously a molecule with polar bonds can be polar in itself. It's like saying I am an atheltic person who can just reach the basketball rim with my head and also I can dunk.

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When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

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Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

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c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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