Answer:
-196 kJ
Explanation:
By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ
-------------------------------------------------------------
2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)
Simplifing the compounds that are in both sides (bolded):
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ
Answer:
[O₃]= 8.84x10⁻⁷M
Explanation:
<u>The photodissociation of ozone by UV light is given by:</u>
O₃ + hν → O₂ + O (1)
<u>The first-order reaction of the equation (1) is:</u>
(2)
<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>
<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>
(3)
<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>
We can calculate the initial ozone concentration using equation (3):
So, the ozone concentration after 10 days is 8.84x10⁻⁷M.
I hope it helps you!
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
Dimension analysis is to be used to solve this problem. First convert 1L to milliliters. That is equivalent to 1000 ml. Then by dimension analysis, multiply the volume ( 1000ml) to the density of oil (0.92 g/ml) resulting to the answer: 920 grams.