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ahrayia [7]
3 years ago
12

When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have th

e greatest effect on the pH of the solution is:
Chemistry
1 answer:
Svetllana [295]3 years ago
5 0

Answer and explanation:

Initially, the pH of the solution will be determined by the dissociation of the weak acid.

Usually, this kind of solution consists of a mixture of weak acid that has not yet reacted and the salt that will be formed by the reaction of the weak acid with a strong base that has been added.

After the dissociation process, the pH will be more affected due to the excess of base in the solution.

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Answer:

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Explanation:

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Do step 3 as outlined in the lab guide. Record your results in the appropriate blanks.
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Explanation:

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3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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