Answer : The correct option is, (C) 26.5 L
Explanation :
The combustion of ethylene is:
![C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)](https://tex.z-dn.net/?f=C_2H_4%28g%29%2B3O_2%28g%29%5Crightarrow%202CO_2%28g%29%2B2H_2O%28g%29)
First we have to calculate the number of moles of water vapor.
From the balanced chemical reaction, we conclude that:
As, 1 mole of
react to give 2 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 0.535 mole of
react to give
moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Now we have to calculate the volume of water vapor.
Using ideal gas equation:
![PV=nRT\\\\V=\frac{nRT}{P}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5CV%3D%5Cfrac%7BnRT%7D%7BP%7D)
where,
P = pressure of gas = 100 kPa
V = volume of gas = ?
T = temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
R = gas constant = 8.314 L.kPa/K.mol
n = number of moles of gas = 1.07 mol
Now put all the given values in the above formula, we get:
![V=\frac{nRT}{P}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D)
![V=\frac{1.07 mol\times 8.314 L.kPa/K.mol\times 298K}{100kPa}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1.07%20mol%5Ctimes%208.314%20L.kPa%2FK.mol%5Ctimes%20298K%7D%7B100kPa%7D)
![V=26.5L](https://tex.z-dn.net/?f=V%3D26.5L)
Therefore, the volume of water vapor is, 26.5 L.
at 60ºC:
106 g KNO₃ -------------- 100 g (H₂O)
? g KNO₃ ------------------ 50.0 g (H₂O)
mass KNO₃ = 50.0 * 106 / 100
mass KNO₃ = 5300 / 100
<span>= </span>53 g of KNO₃
answer (2)
<span>hope this helps!</span>
The lab report contains the theory, process, data and calculation of the experiment. The theory and process are remains fixed for a particular experiment. Thus there is no chance to get error from these two part. The calculation depends upon the reading or data of the experiment. The calculation is also unique and based on the data. Thus the error come from the data of the experiment. As for example for a titration experiment the data recorded in the process from the burret is the source of error, on which the calculation depends.
Answer:
to increase the activation energy of the reaction..
I hope this helps