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german
3 years ago
8

A 35 L sample of N2 is at 65 mmHg. What volume would the gas occupy at 74 mmHg? *

Chemistry
1 answer:
kvv77 [185]3 years ago
8 0

Answer:

About 40 L

Explanation:

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Acid &amp; Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
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<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

  • If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
  • The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

[H_3O^+]=5.012\times 10^{-5}M

The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

  • <u>For 6:</u>

We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0
3 years ago
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