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n= 0,1 mol/ L x0,25 L
=0,025 mol
mr KOH= 56 g/mol
m= 56 g/mol x 0,025 mol
=1,4 g
no 2 ad molar yg hilang?
3.mnhtg mol masing" senyawa
mol BaBr2 = 1 mol/ Lx 0,05 L
= 0,05 mol
BaBr2 --> Ba2+ + 2Br-
0,05 mol 0,1 mol
mol NaBr= 0,4 mol /L x 0,1 L
= 0,04 mol
NaBr -> Na+ + Br-
0,04 mol 0,04 mol