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KonstantinChe [14]
3 years ago
13

A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. what is the partial pressure of oxygen

at 101.4 kpa if phe = 82.5 kpa and pco2 = 0.4 kpa?
Chemistry
2 answers:
zysi [14]3 years ago
8 0

Answer:

18.5 kPa is the partial pressure of oxygen.

Explanation:

Total pressure of the gases = p =  101.4 kPa

Partial pressure of the helium gas = p_{He}=82.5 kPa

Partial pressure of the carbon dioxide gas = p_{CO_2}=0.4 kPa

Partial pressure of the oxygen gas = p_{O_2}=?

Applying Dalton's law of partial pressure:

p=p_{He}+p_{CO_2}+p_{O_2}

101.4 kPa=82.5 kPa+0.4 kPa+p_{O_2}

p_{O_2}=101.4 kPa-82.5 kPa-0.4 kPa=18.5 kPa

18.5 kPa is the partial pressure of oxygen.

pentagon [3]3 years ago
7 0
Pressure of the Mixture = 101.4 kpa  
Pressure of the helium phe = 82.5 kpa 
 Pressure of CO2 pco2 = 0.4 kpa.
 Partial pressure of oxygen = Pressure of the Mixutre - (Pressure of the
helium + Pressure of CO2)
 Partial pressure of oxygen = 101.4 - 82.5 - 0.4
 Partial pressure of oxygen = 18.5 kPa
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emmasim [6.3K]

Answer: The value of the equilibrium constant, Kc, for the reaction is 0.013

Explanation:

Initial concentration of NH_3 = 3.60 M

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The given balanced equilibrium reaction is,

            4NH_3(g)+7O_2(g)\rightleftharpoons 2N_2O_4(g)+6H_2O(g)

Initial conc.       3.60 M        3.60 M              0 M        0 M

At eqm. conc.     (3.60-4x) M   (3.60-7x) M   (2x) M      (6x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[N_2O_4]^2\times [H_2O]^6}{[NH_3]^4\times[Cl_2]}

[N_2O_4]=0.60M

2x = 0.60 M

x= 0.30 M

Now put all the given values in this expression, we get :

K_c=\frac{(2\times 0.30)^2\times (6\times 0.30)^6}{(3.60-3\times 0.30)^4\times (3.60-7\times 0.30)^7}

K_c=0.013

6 0
3 years ago
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DanielleElmas [232]

Answer:  Its "C"

Explanation:

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3 0
3 years ago
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Arte-miy333 [17]

Answer:

7.6 days

Explanation:

Radon is a radioactive element and Radon-222 is it's most stable isotope. The half-life of Radon-222 has been found to be approximately 3.8 days.

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We will use the following relation for calculating time elapsed in the decay

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We can write is as,

(\frac{1}{2} )^2=(\frac{1}{2} )^\frac{t}{3.8}

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5 0
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melamori03 [73]
D) The same number of each type of atom will always be present before and after a chemical reaction takes place.

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Answer:

I know 2

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