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kenny6666 [7]
3 years ago
14

A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of

space and have a pressure of 168 kPa?
Chemistry
1 answer:
Firdavs [7]3 years ago
4 0

Answer: 318 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 231 kPa

P_2 = final pressure of gas = 168 kPa

V_1 = initial volume of gas = 3.25 L

V_2 = final volume of gas = 4.35 L

T_1 = initial temperature of gas = 54^oC=273+54=327K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{231\times 3.25}{327}=\frac{168\times 4.35}{T_2}

T_2=318K

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa

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Which statments regarding the henderson-hasselbalch equation are true?
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Complete question is;

Which statements regarding the Henderson-Hasselbalch equation are true?

1. If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined.

2. At pH = pKa for an acid, [conjugate base] = [acid] in solution.

3. At pH > pKa for an acid, the acid will be mostly ionized.

4. At pH < pKa for an acid, the acid will be mostly ionized.

A. All of the listed statements are true. B. 1, 2, and 3 are true.

C. 2, 3, and 4 are true.

D. 1, 2, and 4 are true.

Answer:

B. 1, 2, and 3 are true.

Explanation:

The formula for the Henderson-Hasselbalch equation is:

pH = pka + log₁₀([A^(-)]/[HA])

Where;

PH is acidity of solution

ka is acid dissociation constant

A^(-) is concentration of conjugate base

HA is concentration of Acid

- For statement 1; If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined;

pH = pka + log₁₀([A^(-)]/[HA])

pH - pka = log₁₀([A^(-)]/[HA])

10^(pH - pka) = ([A^(-)]/[HA])

Since we can find the ratio as seen, then the statement is true

- For statement 2: At pH = pKa for an acid, [conjugate base] = [acid] in solution;

We will substitute pH for pKa;

pH = pH + log₁₀([A^(-)]/[HA])

This give;

0 = log₁₀([A^(-)]/[HA])

10^(0) = [A^(-)]/[HA]

1 = [A^(-)]/[HA]

Thus; [A^(-)] = [HA]

Thus, the statement is true

- For statement 3: At pH > pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is greater than 0 and thus;

10^(pH - pKa) is greater than 1.

Thus;

[A^(-)]/[HA] > 1

[A^(-)] > [HA]

So more acid is ionized than base.

So the statement is true.

- For statement 4: At pH < pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is less than 0 and thus;

10^(pH - pKa) is less than 1.

Thus;

[A^(-)]/[HA] < 1

[A^(-)] < [HA]

So we have more base ionized than acid. So statement is false

7 0
2 years ago
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