Since volume and temperature are constant, this means that pressure and <u>number of moles</u> are <u>directly </u>proportional. the sample with the largest <u>number of moles</u> will have the <u>high </u>pressure.
Since, the ideal gas equation is also called ideal gas law. So, according to ideal gas equations,
PV = nRT
- P is pressure of the sample
- T is temperature
- V is volume
- n is the number of moles
- R is universal gas constant
At constant volume and temperature the equation become ,
P ∝ nR
since, R is also constant. So, conclusion of the final equation is
P ∝ n
The number of moles and pressure of the sample is directly proportion. So, on increasing number of moles in the sample , pressure of the sample also increases.
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<h2>Answer : By weighing the costs and benefits of an environmental issue
</h2><h3>
Explanation :</h3>
The law makers usually conduct many studies before a law is imposed. They try to explore many other options available to the current environmental issue and then come to a conclusion to make a law.
They also weigh the cost aspect along with the benefit of the ongoing environmental issue. They try to come up with something which appears to be cost effective and result bearing.
CO2 is carbon dioxide which is most famous for being in gas form so i would figure if it was exposed to freezing temperatures it would turn into a liquid then maybe a solid<span />
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)
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