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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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If an atom has three shell and number of valence electrons are two. Is this element metal
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3 years ago
Read 2 more answers
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135

8 0
3 years ago
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