The partial pressure of oxygen in a sample of air increases if the temperature is increased.
Answer: Option 1
<u>Explanation:
</u>
According to Guy-Lussac's law, at constant volume, pressure exhibited by the gas molecules will be directly proportional to the temperature of the gas molecules. It is also known that pressure of mixture of gas molecules is the sum of partial pressure of each gas molecule in the mixture.
If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.
So, it can also be concluded that temperature increases the oxygen's partial pressure in air increases.
Answer:
is the oxidizing agent
Explanation:
An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)

![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n


n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Answer:
5.25 moles.
Explanation:
The decomposition reaction of NaN₃ is as follows :

We need to find how many grams of N₂ produced in the process.
From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.
2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,
3.5 moles of sodium azide decomposes to give
moles of nitrogen gas.
Hence, the number of moles produced is 5.25 moles.