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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:

Explanation:

1) Acetic acid

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[H+] = 10-2.96

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Initial 0.103 0 0

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equlibrium 0.103 -x x x

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the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

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pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

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0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

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3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

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neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

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HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

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moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

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PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

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x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

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