Answer:
Explanation:
The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:
Σ
The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:
Solve for Po2:
Thus, the partial pressure of diatomic oxygen is 177.707 torr.
<u><em>If you liked this solution, hit Thanks or give a Rating!</em></u>
According to Boyle's Law, P1V1 = P2V2
where P1 and V1 are initial pressure and volume respectively. P2 and V2 are final pressure and volume receptively.
Given: P2 = 4 P1 and V1 = 10.0l
∴ V2 = 2.5 l
Answer: Final volume of system is 2.5 l
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
Atomic radii increase when going down a group and decreases when going towards the anion periods. So A and D.
I hope you understood
hit me up if you have any other questions :>
