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 general equation for radioactive decay is;
N = N₀e^(-λt)
x - decay constant (λ) - rate of decay
t- time
N - amount remaining after t days , since we are calculating the half life, amount of time it takes for the substance to to be half its original value, its N₀/2
N₀ - amount initially present
substituting the values
N₀/2 = N₀e^(-0.081t)
0.5 = e^(-0.081t)
ln (0.5) = -0.081t
-0.693 = -0.081t
t = 0.693 / 0.081
= 8.55
half life of substance is 8.55 days
Answer:
9.96*10^21
Explanation:
Molar mass of K2O=29*2+16
= 74g per mol
number of moles in the sample= 1.224/ 74
=0.1654
Number of particles in 1 mole=6.0221409*10^23
Number of particles= 0.01654*6.0221409*10^23
=9.96*10^21
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