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3 years ago

In a mixture of He, Ne, and Kr the total pressure is 125 kPa. The partial

Chemistry

1 answer:

tatiyna3 years ago

7 0

Answer:

Partial pressure of He = 73 kPa

Explanation:

Given:

Total pressure = 125 kPa

Partial pressure of Ne = 31 kPa

Partial pressure of Kr = 21 kPa

Find:

Partial pressure of He

Computation:

Total pressure = Partial pressure of Ne + Partial pressure of Kr + Partial pressure of He

125 kPa = 31 kPa + 21 kPa + Partial pressure of He

Partial pressure of He = 73 kPa

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The equation to show the the correct form to show the standard molar enthalpy of formation:

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The equation to show the the correct form to show the standard molar enthalpy of formation:

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3 years ago

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2 years ago

If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce

Natali5045456 [20]

2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

=0.0282669621 g of O2 left over

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