Answer:
4.43 g
Explanation:
The reaction between sodium chloride and flourine gas is given as;
NaCl + F2 --> NaF + Cl2
From the stochiometry of the equation;
1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2
Mass of 1 mol of F2 = 38g
Mass of 1 mol of sodium flouride, NaF = 42g
This means 38g of flourine reacted with NaCl to form 42g of NaF
xg of F2 would form 4.9g of NaF
38 = 42
x = 4.9
x = 4.9 * 38 / 42
x = 4.43 g
Heavy rainfall because that’s a natural thing that happens and can never stop
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6