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xxTIMURxx [149]
3 years ago
11

What is the ph of pure water at 40.0°c if the kw at this temperature is 2.92 × 10-14?

Chemistry
1 answer:
olganol [36]3 years ago
5 0

Answer: pH = 6.77

Explanation:

1) <u>Chemical equilibrium</u>

  • 2 H₂O (l) ⇄ H₃O⁺ (aq) + OH⁻ (aq)

2) <u>Equilibrium constant, Kw</u>

  • Kw = [H₃O⁺] × [OH⁻]
  • By stoichiometry [H₃O⁺] = [OH⁻]. Call it x
  • Kw = x²
  • x² = 2.92 × 10⁻¹⁴ M²
  • x = √ (2.92 × 10⁻¹⁴) = 1.709 × 10⁻⁷ M = [H₃O⁺]

3)<u> pH</u>

  • pH = - log [H₃O⁺] = - log (1.709 × 10⁻⁷) = 6.77
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Answer:

As proteins vary widely in structures than that of nucleic acid there will be greater structural  and functional diversity of proteins  than nucleic acid.

Explanation:

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3 years ago
Chemistry
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Explanation:

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A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

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