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!!!
Answer:
g HCN = MW HCN g/mole HCN * [8.00E19 molecules/6.022E23 molecules/mole HCN] = ??
EmForm = CaxPy
moles Ca = 0.66g / atwt Ca g/mole = ??
moles P = (1.00-0.66)g / atwt P g/mole = ??
x/y = moles Ca / moles P = ?? (express as ratio of two numbers; if both are not intergers multiply each by the same number that make them integers. This ratio gives the empirical formula.
Ca3P2
Plug and SOLVE
<span>Even surfaces that seem smooth have different bumps that you could see with a microscope, so friction occurs even with smooth-looking objects.</span>
Answer:
The molarity of this solution is 14.82 mol/dm3 or 14.82 mol/L
Explanation:
- Molarity is the number of mole present in 1 Litre of solution. Molarity of a solution is a term referred to as concentration of a solution. The unit of Molarity is Mol/dm3 or Mol/L.
- let us make an assumption that the volume of the ammonia solution is 1L or 1dm3. Also, 1L = 1000 mL.
Step 1: calculate the mass of the solution
Density = 0.90g/ml (from the question)
Density = mass/ volume
Therefore Mass = density x volume
= 0.90g/ml x 1000ml
mass = 900 g
Step 2: calculate the mass of NH3 present in the solution
Since the concentrated aqueous of ammonia is 28%, It signifies that 1000ml of the solution contains 28% Ammonia
Recall from the above calculation that the mass of 1000 ml of solution is 900 g.
Therefore the mass of ammonia will be 28% of 900 g
mass of NH3 = 0.28 x 900 g
= 252 g
Step 3: calculate the number of mole of NH3
mole = mass/ molar mass
molar mass of NH3 = 17 g/mol
Therefore mole of NH3 = 252/17
= 14.82 mol
Step 4: Calculate Molarity
Molarity = number of moles/ volume of solution in Litre (L)
Molarity = 14.82 / 1
Molarity = 14.82 mol/L
Answer:
a
Explanation:
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