The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
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Argon = Ar
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Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is
= 283.725 kJ ⋅ mol − 1
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Answer : The pressure in the flask after reaction complete is, 2.4 atm
Explanation :
To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

where,
P = final pressure in the flask = ?
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 
V = volume = 4.0 L
= moles of
= 0.20 mol
= moles of
= 0.20 mol
Now put all the given values in the above expression, we get:


Thus, the pressure in the flask after reaction complete is, 2.4 atm