Convert mole to gram by multiplying the molar mass of sodium
0.500mol Na x 22.990g = 11.495g of Na
a) The total pressure of the system is 1.79 atm
b) The mole fraction and partial pressure of hydrogen is 0.89 and 1.59 atm respectively
c) The mole fraction and the partial pressure of argon is 0.11 and 0.19 atm.
<h3>What is the total pressure?</h3>
We know tat we can be able to obtain the total pressure in the system by the use of the ideal gas equation. We would have from the equation;
PV = nRT
P = pressure
V = volume
n = Number of moles
R = gas constant
T = temperature
Number of moles of hydrogen = 14.2 g/2g = 7.1 moles
Number of moles of Argon = 36.7 g/40 g/mol
= 0.92 moles
Total number of moles = 7.1 moles + 0.92 moles = 8.02 moles
Then;
P = nRT/V
P = 8.02 * 0.082 * 273/100
P = 1.79 atm
Mole fraction of hydrogen = 7.1/8.02 = 0.89
Partial pressure of hydrogen = 0.89 * 1.79 atm
= 1.59 atm
Mole fraction of argon = 0.92 / 8.02
= 0.11
Partial pressure of argon = 0.11 * 1.79 atm
= 0.19 atm
Learn more about partial pressure:brainly.com/question/13199169
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In order to answer this question, the units of volume must be consistent. In this problem, we decide the unit m3 to be uniform. Option A is equal to 12 m3, option b is equal to 1.2x10^8/100^3 or 120 m3. Option C is 2.0 x10^4/ 10^3 or 20 m3. Option D is 1.2x10^8/ 1000^3 or 0.12 m3. The greatest volume is option b. 120 m3.
The water will boil at C) 80°C
<h3>Further explanation</h3>
Given
Vapour pressure of water = 47 kPa
Required
Boiling point of water
Solution
We can use the Clausius-Clapeyron equation :
Vapour pressure of water at boiling point 100°C=101.325 kPa
ΔH vap for water at 100°C=40657 J/mol
R = 8.314 J/mol K
T₁=boiling point of water at 101.325 kPa = 100+273=373 K
Input given values :
Air is a homogeneous mixture, consisting of <span>nitrogen, oxygen, carbon dioxide and other gases present in air.</span>
