A gas has a pressure of 2.31 atm and occupies
1 answer:
Answer:
The pressure of the gas after getting compressed is 1.92 L.
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
(At constant temperature)
where,
are initial pressure and volume.
are final pressure and volume.
We are given:
Putting values in above equation, we get:
The pressure of the gas after getting compressed is 1.92 L.
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Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is
Explanation:
The chemical equation for this decomposition of ammonia is
↔
The initial concentration of ammonia is mathematically represented a
The initial concentration of nitrogen gas is mathematically represented a
So looking at the equation
Initially (Before reaction)
During reaction(this is gotten from the reaction equation )
(this implies that it losses two moles of concentration )
(this implies that it gains 1 moles)
(this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
Now since
Now the equilibrium constant is
substituting values
