Answer:
Kc for this reaction is 0.06825
Explanation:
Step 1: Data given
Number of moles formaldehyde CH2O = 0.055 moles
Volume = 500 mL = 0.500 L
At equilibrium, the CH2O(g) concentration = 0.051 mol
Step 2: The balanced equation
CH2O <=> H2 + CO
Step 3: Calculate the initial concentrations
Concentration = moles / volume
[CH2O] = 0.055 moles . 0.500 L
[CH2O] = 0.11 M
[H2] = 0M
[CO] = 0M
Step 4: The concentration at the equilibrium
[CH2O] = 0.11 - X M = 0.051 M
[H2] = XM
[CO] = XM
[CH2O] = 0.11 - X M = 0.051 M
X = 0.11 - 0.051 = 0.059
[H2] = XM = 0.059 M
[CO] = XM = 0.059 M
Step 5: Calculate Kc
Kc = [H2][CO]/[CHO]
Kc = (0.059 * 0.059) / 0.051
Kc = 0.06825
Kc for this reaction is 0.06825
The mechanical energy is <span>98 J</span>
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O