Question 1 a sample of 0.255 mole of gas has a volume of 748 ml at 28°c. calculate the pressure of this gas. (r= 0.0821 l ∙ atm
2 answers:
Answer : The pressure of the gas is, 8.42 atm
Explanation :
Using ideal gas equation,
where,
P = pressure of the gas = ?
V = volume of the gas = 748 ml = 0.748 L
conversion used : (1 L = 1000 ml)
T = temperature of the gas =
n = number of moles of the gas = 0.255 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get the pressure of the gas.
Therefore, the pressure of the gas is, 8.42 atm
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Explanation:
The given data is as follows.
= 85.0 ml,
= 0.9 M
= 85.0 ml,
= 0.9 M
Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.
So, No. of moles = Molarity × Volume
= (as 1 L = 1000 ml so, 85 ml = 0.085 L)
= 0.076 mol
As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.
= 4.271 kJ
or, = 4271 J (as 1 kJ = 1000 J)
Therefore, heat released = - heat of gained by calorimeter
Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.
Hence, mass of HCl = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Similarly, mass of KOH = = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Hence, total mass of the solution = 85 g + 85 g
= 170 g
Also, q =
4271 J =
0.0773 =
=
Thus, we can conclude that final temperature of the mixed solution is .