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kotegsom [21]
3 years ago
13

Caffeine (C_8H_10N_4O_2) is a weak base with a K_b value of 4 times 10^-4. The pH of a 0.01 M solution of caffeine is in the ran

ge of: a. 2-3 b. 5-6 c. 7-8 d. 9-10 e. 11-12
Chemistry
1 answer:
docker41 [41]3 years ago
3 0

Answer:

The pH of the solution lies from 11 to 12.Hence, option e is correct.

Explanation:

The value of K_b for caffine = 4\times 10^{-4}

CafOH(aq)\rightleftharpoons Caf(aq)+OH^-(aq)

Initial

   0           0.01 M       0

AT equilibrium:

  x          (0.01 -x)M      x

K_b=\frac{x(0.01-x)}{(x)}

4\times 10^{-4}=\frac{x(0.01-x)}{(x)}

Solving for x:

x = 0.0096 M

The pOH of the solution is given by :

pOH=-\log[OH^-}

pOH=-\log[x]

pOH=-\log[0.0096]

pOH = 2.02

pH= 14 - pOH = 14 - 2.02 = 11.98

The pH of the solution lies from 11 to 12.

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Write the number 345 in scientific notation, with the same number of significant figures:
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3.45 x 10^2

Explanation:

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24
Setler79 [48]

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

To calculate the partial pressure of helium, we use the formula:

P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}

Putting values in above equation, we get:

745=245+119+163+p_{He}\\p_{He}=218mmHg

Hence, the partial pressure of Helium is 218 mmHg.

  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

6 0
3 years ago
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