From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
Learn more about stoichiometry: brainly.com/question/13110055
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(1) 14C also known as Carbon-14, is used to radio-date organic materials as well as earth materials.
U must burn it
or
Make it under go chemical treatment
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
