Question 1 :
V1/T1 = V2/T2
3.0L/273K = V2/373K
To get the value of Z, cross multiply
3.0L x 373K = 273K x V2
1119 = 273V2
Divide both sides by 273
1119/273 = 273V2/273
4.10L = V2
The new volume is 4.10 liters
Question 2 :
P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Question 3 :
Given that:
Volume of gas V = 4.80L
(since 1 liter = 1dm3
4.80L = 4.80dm3)
Temperature T = 62°C
Convert Celsius to Kelvin
(62°C + 273 = 335K)
Pressure P = 2.9 atm
Number of moles of gas N = ?
Apply ideal gas equation
pV = nRT
2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)
13.92 atm dm3 = nx 2.747 atm dm3 mol-1
n = 13.92/2.747
n = 5.08 moles
There are 5.08 moles of gas contained in the sample
Question 4 :
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
The pressure of the gas is 64.6 mm hg
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The heat of solution is -51.8 kJ/mol
<h3>What is the heat of solution?</h3>
We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.
Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles
Temperature rise = 26.0 ∘c
Mass of the water = 100.0 grams
Heat capacity = 4.184 j/g⋅°c
Then;
ΔH = mcθ
ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ
Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol
Learn more about heat of solution:brainly.com/question/24243878
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The answer is B. Fungi,Protists
