Match each statements with the state of matter it describes?
1 answer:
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19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.
<h3>What is vapour pressure?</h3>Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2
Volume of (V) = 40 liter
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm
Moles of (n) can be calculated by ideal gas equation.
PV = nRT
n = 1.007 40 ÷ 0.0821 295 = 1.663
Balanced chemical reaction;
2 + 13
---> 8
+ 10
From reaction;
13 moles require 2 moles
So, 1.663 moles will require = 2 x 1.663 ÷13 = 0.256 moles of
Thus is a limiting reagent. So it will drive the yield of
.
Moles of produced = (8/2) 0.2 = 0.8 moles
Pressure of (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm
Applying the ideal gas equation for ,
PV = nRT
0.985 V = 0.8 0.0821 x 295
V = 19.7 liter
The volume of produced = 19.7 liter.
Learn more about the vapour pressure here:
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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.
Calculations:
The second order rate equation is:
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