Answer:
a) 3.98 x 10^-10
Explanation:
Hello,
In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Hence, solving for the concentration of hydronium:
![[H^+]=10^{-pH}=10^{-9.40}\\](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-9.40%7D%5C%5C)
![[H^+]=3.98x10^{-10}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D3.98x10%5E%7B-10%7DM)
Therefore, answer is a) 3.98 x 10^-10
Best regards.
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer : The number of moles of solute 
is, 0.0788 moles.
Explanation : Given,
Molarity = 0.225 M
Volume of solution = 0.350 L
Formula used:
Now put all the given values in this formula, we get:
Therefore, the number of moles of solute is, 0.0788 moles.
Answer:
A). 92.02g
Explanation:
Equation of the reaction;
N2 (g)+ 2O2(g)------> 2NO2(g)
Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved.
Reaction of one mole of nitrogen gas with two moles of oxygen gas yields two moles of nitrogen dioxide.
Mass of two moles of nitrogen dioxide= 2[14 + 2(16)] = 2[14+32]= 2[46]= 92 gmol-1
Therefore; Mass of two moles of nitrogen dioxide is 92
