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Xelga [282]
3 years ago
12

He partial pressure of oxygen gas in our atmosphere is 0.21 atm. this is the partial pressure at which human lungs have evolved

to be able to breathe this gas. a scuba diver, will thus still have to breath oxygen at this pressure even when diving way down in the water. if a mixture of helium and oxygen (heliox) in his tank is at a pressure of 8.30 atm, what must the partial pressure be of helium to keep the partial pressure of oxygen at 0.21 atm?
Chemistry
1 answer:
solong [7]3 years ago
6 0
Hello!

The partial pressure of helium to keep the partial pressure of oxygen at 0,21 atm in a scuba-diver tank is 8,09 atm

To solve this question, we can use the Dalton's Law, which states that the total pressure in a container with a mixture of gases is the sum of the partial pressures o each individual gas. For the case of this mixture the Dalton's Law is as follows:

P_{tot}=P_{He} +P_{O_2}

In this equation, we need to clear for PHe, knowing that the PO₂ should be 0,21 atm, to find the required pressure of Helium:

P_{He}=P_{tot} -P_{O_2}=8,30atm-0,21atm=  8,09 atm

Have a nice day!
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Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
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Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

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\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

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Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

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V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

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Hey there!:

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