Answer: The correct option is (c). The total pressure doubles.
Solution:
Initially, only 4 moles of oxygen gas were present in the flask.
(
) ( according to Dalton's law of partial pressure)
....(1)
= Total pressure when only oxygen gas was present.
Final total pressure when 4 moles of helium gas were added:
partial pressure of oxygen in the mixture :
Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.
= Total pressure of the mixture.
from (1)
On rearranging, we get:
The new total pressure will be twice of initial total pressure.
A metalloid is a metal and a nonmetal
Answer:
Final pressure in (atm) (P1) = 6.642 atm
Explanation:
Given:
Initial volume of gas (V) = 12.5 L
Pressure (P) = 784 torr
Temperature (T) = 295 K
Final volume (V1) = 2.04 L
Final temperature (T1) = 310 K
Find:
Final pressure in (atm) (P1) = ?
Computation:
According to combine gas law method:
⇒ Final pressure (P1) = 5,048.18877 torr
⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760
⇒ Final pressure in (atm) (P1) = 6.642 atm
Answer:
12.62 L
Explanation:
First, we have to calculate the moles corresponding to 18.0 g of oxygen gas (MW 32.0).
18.0 g × (1 mol/32.0 g) = 0.563 mol
Then, we can find the volume occupied by 0.563 moles of oxygen at STP (273,15 K, 1.00 atm) using the ideal gas law.
P × V = n × R × T
V = n × R × T / P
V = 0.563 mol × 0.0821 atm.L/mol.K × 273.15 K / 1.00 atm
V = 12.62 L
