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AURORKA [14]
3 years ago
7

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

total pressure of 2.07 x 104 kPa, then what is the partial pressure of helium?
Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Answer: 1.61 x 10⁴ kPa

Dalton's law <u>states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture.</u> According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction <u>is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt)</u>, which is calculated by adding the moles of all its components:

xi = \frac{n_{i} }{n_{t} }

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ =  10 g x \frac{1 mol}{31.998 g} = 0.313 mol

n₂ =  50 g x \frac{1 mol}{8.005 g} = 6.246 mol

n₃ =  40 g x \frac{1 mol}{28.013 g} = 1.428 mol

Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ =  \frac{6.246 mol}{7.987 mol} = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

So, <u>the partial pressure of helium if a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa is  1.61 x 10⁴ kPa</u>

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Answer:

1. pH = 1.23.

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Explanation:

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1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

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pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

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