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Nutka1998 [239]
3 years ago
10

A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container. If the total pressure is 1.24

atm. What is the partial pressure of hydrogen? Use mc032-1.jpg.
Chemistry
1 answer:
Degger [83]3 years ago
6 0
  The   partial  pressure of hydrogen is 0.31  atm

calculation

find the number of  hydrogen   moles the container, that is

25/100  x 6.4  =1.6 moles of hydrogen

find the  partial pressure for hydrogen  in 1.6 moles

that is   6.4  moles=  1.24 atm
            1.6  moles= ?

by  cross  multiplication

1.6moles  x1.24  atm/ 6.4 moles=  0.31 atm
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<h3>Further explanation</h3>

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Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h
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Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

<u>Step 1:</u> Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

<u>Step 2:</u> Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

<u>Step 3:</u> Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

<u> </u>

<u>Step 4</u>: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

<u>Step 5:</u> Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0
3 years ago
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