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olga2289 [7]
2 years ago
9

How many grams of oxygen gas occupy 12.3 L of space at 109.4 kPa and 15.4oC?

Chemistry
1 answer:
I am Lyosha [343]2 years ago
3 0

17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of the gas must be calculated first as follows:

PV = nRT

Where;

  • P = pressure = 1.0796941atm
  • V = volume = 12.3L
  • n = number of moles
  • T = temperature = 288.4K
  • R = gas law constant = 0.0821 Latm/molK

1.079 × 12.3 = n × 0.0821 × 288.4

13.27 = 23.68n

n = 13.27/23.68

n = 0.56mol

Mass = 0.56 × 32

mass of oxygen gas = 17.93g

Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.

Learn more about mass at: brainly.com/question/19694949

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Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

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Step 3: The initial concentration

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[I2] = 0.453 M

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Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

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[HI] = 2X = 2*0.3563 = 0.7126 M

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3 years ago
How many moles of CCI are there in 78.2 g of CCI.?
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Answer:

0.508 mole

Explanation:

NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.

The number of mole present in 78.2 g of CCl₄ can be obtained as follow:

Mass of CCl₄ = 78.2 g

Molar mass of CCl₄ = 12 + (35.5×4)

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Mole of CCl₄ =?

Mole = mass / molar mass

Mole of CCl₄ = 78.2 / 154

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