Answer:
Option C. Triple the number of moles
Explanation:
From the ideal gas equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of mole
R is the gas constant
T is the absolute temperature.
Making V the subject of the above equation, we have:
PV = nRT
Divide both side by P
V = nRT / P
Thus, we can say that the volume (V) is directly proportional to both the number of mole (n) and absolute temperature (T) and inversely proportional to the pressure (P). This implies that and increase in either the number of mole, the absolute temperature and a decrease in the presence will cause the volume to increase.
Thus, the correct option is option C triple the number of moles. This can further be seen as illustrated below:
Initial volume (V1) = 12 L
Initial mole (n1) = 0.5 mole
Final mole (n2) = triple the initial mole = 3 × 0.5 = 1.5 mole
Final volume (V2) =?
From:
V = nRT / P, keeping T and P constant, we have:
V1/n1 = V2/n2
12/0.5 = V2/1.5
24 = V2/1.5
Cross multiply
V2 = 24 × 1.5
V2 = 36 L.
Thus Option C gives the correct answer to the question.
Solar- the sun is not always shining, for example at night you can't get any energy from the sun
Wind-it's not always windy
Water- there can be drought
Really the main issue of all three of these is that they are not always available
Answer:
capacity factor = 0.952
Availability factor = 0.958
Explanation:
1) capacity factor
capacity factor = actual power output / maximum power output
= (actual power output)/(efficiency * rated power output)
= 0.952
2) Availability factor
Availability factor = Actual operation time period/ total time period
= 23/24 = 0.958
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
