Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -

mole fraction of Ne -

mole fraction of Ar -

partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
Answer:
the salt is a solute in this instance
5.2/ 24.3 = 0.214
0.214 x 2 = 0.428
0.428 x 36.5= 15.6g
Answer: The answer is 68142.4 Pa
Explanation:
Given that the initial properties of the cylindrical tank are :
Volume V1= 0.750m3
Temperature T1= 27C
Pressure P1 =7.5*10^3 Pa= 7500Pa
Final properties of the tank after decrease in volume and increase in temperature :
Volume V2 =0.480m3
Temperature T2 = 157C
Pressure P2 =?
Applying the gas law equation (Charles and Boyle's laws combined)
P1V1/T1 = P2V2/T2
(7500 * 0.750)/27 =( P2 * 0.480)/157
P2 =(7500 * 0.750* 157) / (0.480 *27)
P2 = 883125/12.96
P2 = 68142.4Pa
Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa
Answer:
I' really don't know I'm sorry