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Juliette [100K]
3 years ago
8

Can someone pls explain? :'(

Chemistry
1 answer:
telo118 [61]3 years ago
7 0
It is 0.07 g for your answer
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6. At a glance, it appears like the law of
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Energy and mass equivalence

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4 0
3 years ago
What should be the mole fraction of O2 in the gas mixture the diver breathes in order to have the same partial pressure of oxyge
Nutka1998 [239]

The question is incomplete, here is the complete question:

A scuba diver is at a depth of 355 m, where the pressure is 36.5 atm.

What should be the mole fraction of O_2 in the gas mixture the diver breathes in order to have the same partial pressure of oxygen in his lungs as he would at sea level? Note that the mole fraction of oxygen at sea level is 0.209.

<u>Answer:</u> The mole fraction of oxygen in the gas mixture is 0.00573

<u>Explanation:</u>

To calculate the partial pressure, 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 oxygen at sea level = ?

p_T = total pressure at sea level = 1.00 atm

\chi_A = mole fraction of oxygen at sea level = 0.209

Putting values in equation 1, we get:

p_{O_2}=1.00atm\times 0.209\\\\p_{O_2}=0.209atm

As, partial pressure of the oxygen in the diver's lungs is equal to the partial pressure of oxygen at sea level

We are given:

p_T=36.5atm\\p_{O_2}=0.209atm

Putting values in equation 1, we get:

0.209atm=36.5atm\times \chi_{O_2}\\\\\chi_{O_2}=\frac{0.209}{36.5}=0.00573

Hence, the mole fraction of oxygen in the gas mixture is 0.00573

3 0
3 years ago
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