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omeli [17]
1 year ago
7

Oxygen gas generated in the thermal decomposition of potassium chlorate is collected over water. The volume of gas collected is

0.128 L, at a temperature of 297K and a pressure of 762 mm Hg. The vapour pressure of H2O at 297K is 22.4 torr. What is the mass of oxygen collected?
Chemistry
1 answer:
Leya [2.2K]1 year ago
5 0

The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g

<h3>How to determine the mole of oxygen produced </h3>

We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:

  • Volume (V) = 0.128 L
  • Temperature (T) = 297 K
  • Pressure (P) = 762 – 22.4 = 739.6 mmHg
  • Gas constant (R) = 62.363 mmHg.L/Kmol
  • Number of mole (n) =?

PV = nRT

739.6 × 0.128 = n × 62.363 × 297

Divide both sides by 62.363 × 297

n = (739.6 × 0.128) / (62.363 × 297)

n = 0.0051 mole

Thus, the number of mole of oxygen gas produced is 0.0051 mole

<h3>How to determine the mass of oxygen collected</h3>

Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'

  • Mole = 0.0051 mole
  • Molar mass of oxygen gas = 32 g/mole
  • Mass of oxygen =?

Mole = mass / molar mass

0.0051 = mass of oxygen / 32

Cross multiply

Mass of oxygen = 0.0051 × 32

Mass of oxygen = 0.16 g

Thus, we can conclude that the mass of oxygen gas collected is 0.16 g

Learn more about ideal gas equation:

brainly.com/question/4147359

#SPJ1

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So, consider the present weight in the percentage of ^{235}U to be  y%

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In( \dfrac{N_1}{N_o}) = - \lambda t_1

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In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

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Now, equating  equation (1) and (2) together, we have:

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\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

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