Asked by Cutie
Calculate the amount concentration of the cations in the following solutions:
A) 0.5 mol/L sodium carbonate
B) 0.2 mol/L ammonium sulfite
C) 1.5 mol/L iron 3 sulfate
A) 0.5 mol/L sodium carbonate
B) 0.2 mol/L ammonium sulfite
C) 1.5 mol/L iron 3 sulfate
Answers
Answered by
DrBob222
I'll do the middle one.
0.2 mols/L (NH4)2SO3
The cation is NH4^+
For every 1 mol (NH4)2SO3 you have 2 mols NH4^+; therefore, (NH4+) = twice the (NH4)2SO3 and that give 0.2 mols/L x 2 = 0.4 mols/L = 0.4 M.
a is Na2CO3 formula.
c is Fe2(SO4)3 formula.
0.2 mols/L (NH4)2SO3
The cation is NH4^+
For every 1 mol (NH4)2SO3 you have 2 mols NH4^+; therefore, (NH4+) = twice the (NH4)2SO3 and that give 0.2 mols/L x 2 = 0.4 mols/L = 0.4 M.
a is Na2CO3 formula.
c is Fe2(SO4)3 formula.
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