The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
The mass of the solid product of the reaction is determined by weighing.
Answer:
c) NOx and VOC
oxides of nitrogen (NOx) and volatile organic compounds (VOC)
Explanation:
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Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
