Answer:
![[H^+]=1.78x10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.78x10%5E%7B-8%7DM)
Explanation:
Hello there!
In this case, according to the given information about the pH, it is firstly necessary for us to remember that the pH is defined as the potential of the hydrogen ions in the solution and the concentration of those ions represents how many of them are present in the solution; in such a way, it is possible for us use:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Whereas the concentration of hydrogen ions can be calculated as follows:
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
So we plug in the given pH to obtain:
![[H^+]=10^{-7.75}=1.78x10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-7.75%7D%3D1.78x10%5E%7B-8%7DM)
Regards!
Need more to the question
The given equation is already balanced. We can use the stoichiometric ratio given by the equation to find how many moles of KNO3 are produced. According to the equation, for every mole of Pb(NO3)2, 2 moles of KNO3 are produced.
It means that 194moles of KNO3 are produced.
In order to determine if a reactant is in excess or limiting, we must first know the required amount of reactants. This is referred to as the stoichometric amount of reactant, and it is obtained from the chemical equation.
From the equation, we form a ratio of the reactants. The reactant supplied in excess of that ratio is the excess reactant, while the other is the limiting reactant.
Answer: 176 g N2O, 72 g H2O
Explanation:
1. 64g O2 * (1 mol O2)/(32 g O2) * (2 mol N2O)/(1 mol O2) * (44g N2O)/(1 mol N2O) = 176 g N2O
2. 32g CH4 * ( 1 mol CH4)/(16 g CH4) * (2 mol H2O)/(1 mol CH4) * (18 g H2O)/(1 mol H2O) = 72 g H2O
