BRAINIEST AND MAX POINTS
2 answers:
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Answer:
32.7 grams of Zn will remained in the crucible after cooling.
Explanation:
..[1]
..[2]
Adding [1] + 2 × [2] we get:
..[3]
Moles of ZnS in crucible = 0.50 mol
According to reaction [3]. 2 moles of ZnS gives 2 moles of Zn.
Then 0.50 moles of ZnS will give:
of Zn.
Mass of 0.50 moles of Zn =
= 0.50 mol × 65.4 g/mol =32.7 g
32.7 grams of Zn will remained in the crucible after cooling.
Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J
Explanation :
First we have to calculate the value of .
As,
So,
Given :
Now we have to calculate the initial volume of gas.
Formula used :
where,
= initial pressure of gas = 4.25 atm
= initial volume of gas = ?
= initial temperature of gas = 300 K
n = number of moles of gas = 1.0 mol
R = gas constant = 0.0821 L.atm/mol.K
Now we have to calculate the final volume of gas by using reversible adiabatic expansion.
where,
= initial pressure of gas = 4.25 atm
= final pressure of gas = 2.50 atm
= initial volume of gas = 5.80 L
= final volume of gas = ?
= 1.4
Now put all the given values in above formula, we get:
Now we have to calculate the final temperature of gas.
Formula used :
where,
= final pressure of gas = 2.50 atm
= final volume of gas = 8.47 L
= final temperature of gas = ?
n = number of moles of gas = 1.0 mol
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in above formula, we get:
Now we have to calculate the work done.
where,
w = work done = ?
n = number of moles of gas =1.0 mol
= initial temperature of gas = 300 K
= final temperature of gas = 258 K
Now put all the given values in above formula, we get:
Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J