Answer:
Part A
K = (K₂)²
K = (K₃)⁻²
Part B
K = √(Ka/Kb)
Explanation:
Part A
The parent reaction is
2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)
The equilibrium constant is given as
K = [AlBr₃]²/[Al]²[Br₂]³
2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)
K₂ = [AlBr₃]/[Al][Br₂]¹•⁵
It is evident that
K = (K₂)²
3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)
K₃ = [Al][Br₂]¹•⁵/[AlBr₃]
K = (K₃)⁻²
Part B
Parent reaction
S(s) + O₂(g) ⇌ SO₂(g)
K = [SO₂]/[S][O₂]
a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)
Ka = [SO₃]²/[S]²[O₂]³
[SO₃]² = Ka × [S]²[O₂]³
b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Kb = [SO₃]²/[SO₂]²[O₂]
[SO₃]² = Kb × [SO₂]²[O₂]
[SO₃]² = [SO₃]²
Hence,
Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]
(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³
(Ka/Kb) = [SO₂]²/[S]²[O₂]²
(Ka/Kb) = {[SO₂]/[S][O₂]}²
Recall
K = [SO₂]/[S][O₂]
Hence,
(Ka/Kb) = K²
K = √(Ka/Kb)
Hope this Helps!!!
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Answer:
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Answer:
Q1. a) 4Al + 3O₂ ➟ 2Al₂O₃
b) 7.4 moles
c) 11.1 moles
Explanation:
To balance an equation, ensure that the total number of atoms of each element on both sides are equal.
Al + O₂ ➟ Al₂O₃
On the left side of the arrow, you would find the reactants while the product(s) is found on the left hand side.
<u>Reactants</u>
Al atoms: 1
O atoms: 2
<u>Product</u>
Al atoms: 2
O atoms: 3
After balancing,
4Al + 3O₂ ➟ 2Al₂O₃
We have 4 Al atoms and 6 O atoms on both sides.
b) The balanced equation tells us the mole ratio of Al to Al₂O₃.
Al: Al₂O₃
= 4: 2 (÷2 throughout)
= 2: 1
This means that for every 1 mole of Al₂O₃, 2 moles of Al is needed.
Since we need 3.7 moles of Al₂O₃,
number of moles of Al needed
= 2×3.7
= 7.4
c) 4Al + 3O₂ ➟ 2Al₂O₃
For every 4 moles of Al, 3 moles of O are needed.
For each mole of Al,
number of moles of O needed
= 3÷4
= 0.75
For 14.8 moles of Al,
number of moles of O required
= 0.75 ×14.8
= 11.1