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
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Here we have to calculate the amount of 
ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄ 
g of ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
the oxidation states of the elements before and after the reaction is;
Pb oxidation state changes from 0 to +2
SO₄²⁻ ion there's no change in the oxidation state during the reaction
Au oxidation state changes from +3 to 0
reduction reactions are when there's a decrease in the oxidation state of the species
oxidation reactions are when theres an increase in the oxidation state of the species
the element where there's a decrease in oxidation state is Au.
Therefore Au gets reduced.
answer is B) Au
