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dezoksy [38]
3 years ago
7

A sample of helium (He) effuses 2.0 times faster than another gas. What is the molar mass of the other gas?

Chemistry
1 answer:
Mariana [72]3 years ago
6 0

Answer:

The correct answer is 16 gram per mole.

Explanation:

Let A be the gas helium, and B be the unknown gas. It is clearly mentioned in the question that the effusion rate of helium gas is two times more than that of gas B. The molar mass of helium is 4.0 gram per mole. To solve the problem, Grahm's law is used, that is,  

Rate of effusion A/rate of effusion B = √ (Molar mass B/Molar mass of A

2.0 = √Molar mass of B/4.0 gram per mole.  

Now squaring both the sides we get,  

4.0 = Molar mass of B / 4.0

The molar mass of B = 16 gram per mole.  

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The two processes that occur on dissolving a solute in a solvent are dissolution and crystallization and in an unsaturated solution the rate of dissolution is greater than the rate of crystallization

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25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?
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pH = 11.95≈12

Explanation:

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0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

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Kw=[H+]*[OH-]= 10^(-14)

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NaOH -> Na^{+} + OH^{-}

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

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pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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