Answer:
AT THE END OF 80% DISSOLUTION, THE PRESSURE OF NO2 HAS CHANGED FROM 99kPa TO 139.97kPa
Explanation:
P1 = 99 kPa
P2 = unknown
From the reaction,
2 mole of NO2 will produce 2 mole of NO
We can also say that 1 mole of NO2 will produce 1 mole of NO
At 56.6 % of NO2, 0.566 mole of NO2 will be consumed
At STP, 1 mole of a substance will occupy 22.4 dm3 volume
0.566 mole will occupy ( 22.4 * 0.566 / 1) dm3 volume
= 39.58 dm3 volume
V1 = 39.56 dm3
At the new percent of 80%, 0.80 mole of NO2 will be consumed
Since, 1 mole = 22.4 dm3
0.80 mole = (22.4 / 0.80) dm3
= 28 dm3
V2 = 28 dm3
Using the equation of Boyle's law which shows the relationship between pressure and volume of a given mass of gas at constant temperature, we have:
P1 V1 = P2 V2
Re-arranging to make P2 the subject of formula:
P2 = P1V1 / V2
P2 = 99 kPa * 39.56 / 28
P2 = 3916.44 kPa / 28
P2 = 139.87 kPa
So at 80 % dissociation of NO2, the pressure has changed from 99 kPa to 139.97 kPa.
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Answer:
The percent yield of the reaction is 82.5%
Explanation:
Let's work with moles to get the percent yield.
Mass / Molar mass =
652.5 g / 158.03 g/m = 4.13 moles
If the theoretical yield of the reaction is 5 moles but we only made 4.13 moles, the percent yield will be:
(Produced yield / Theoretical yield) . 100 =
(4.13 / 5) . 100 = 82.5 %
The answer is air.
*Couldn't see the last two options*