Answer:
221 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 4.1 L
Initial temperature (T₁) = 25 °C
= 25 °C + 273
= 298 K
Final volume (V₂) = 6.8 L
Final temperature (T₂) =?
The final temperature of the gas can be obtained as follow:
V₁ / T₁ = V₂ / T₂
4.1 / 298 = 6.8 / T₂
Cross multiply
4.1 × T₂ = 298 × 6.8
4.1 × T₂ = 2026.4
Divide both side by 4.1
T₂ = 2026.4 / 4.1
T₂ ≈ 494 K
Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:
°C = K – 273
K = 494
°C = 494 – 273
°C = 221 °C
Thus the final temperature of the gas is 221 °C
The amount of joules of heat that are lost when 150.0 g of steam are cooled from 124 °c to 86 °c is = -11343 joules
calculation
heat(Q) = mass(m) x specific heat capacity(C) x change in temperature (ΔT)
where,
Q=? joules
M=150.0 g
C for steam = 1.99 j/g/°c
ΔT= 86°c-124°c = -38°c
Q is therefore = 150.0 g x 1.99 j/g/°c x -38°c =-11343 joules
Answer:
A) increasing the pressure
Explanation:
Given the exothermic reaction:
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
the question asks what will make the number of moles of SO₃(g) to increase.
When an equilibrium is disturbed, the system will shift in order to counteract the change (see <em>Le Chatelier's Principle</em>) So <em>when the pressure is increased in a reaction involving gases, the equilibrium will shift trying to decrease the number of moles </em>(because pressure is produced by molecules hitting the container), that is in this case it will shift towards the right side, towards the production of SO₃, thus increasing the number of moles of SO₃.
The number of moles in the right side is 2 and the number of moles in the left side is 3.
<span>Actually, the heat of reaction hrxn s calculated by taking
the sum of the heats of formation of the products minus the sum of the heats of
formation of the reactants. However, at heat of formations of pure elements at
atmospheric conditions is zero, therefore the hf of N2 is not important since
it is zero anyway.</span>