Answer:
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of chlorine gas = 
1 atm = 760 Torr
V = Volume of chlorine gas = 385 mL = 0.385 L ( 1 mL - 0.001 L)
n = number of moles of chlorine gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of chlorine gas =25 °C= 25 + 273 K = 300 K
Putting values in above equation, we get:


According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide ,then 0.01573 moles of chlorine gas will be obtained from :
of manganese dioxide
Mass of 0.01573 moles of manganese dioxide:
0.01573 mol × 86.94 g/mol = 1.368 g
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.
The volume of one mole of gas is 22.4 liters at 0 °C and 760 Torr. The volume of the given condition can be calculated using:
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4)/273 = (792 * V₂) / 321.7
V₂ = 25.3 Liters
The Mr of He = 4
The Mr of Ne = 20
Density of He = 4/25.3 = 0.1581 g/L
Density of Ne = 20/25.3 = 0.7905 g/L
Let x be the fraction of Ne
1 - x is the fraction of He
avg density = sum[(mole fraction * Mr)/V]
0.6048 = (20x + 4(1 - x))/25.3
x = 0.70
C. because heat the mixture filter djrjd