A substances an analysis gave the following percentage composition. Na=43.4℅ C=11.3 ℅ O=54.3℅ determine EF and MF.molecular mass = 106u. ( C-12, O-16, Na-23)
Answers
Answered by
1
i do not know sorry .......
Answered by
0
Na 43.4 ÷ 23 = 1.88 2 4
C 11.3 ÷ 12 = 0.94 1 2
O 54.3 ÷ 16 = 3.39 3.5 7
EF = Na4C2O7
MF----
Mass of EF 23×4+12×2+16×7
= 228
EF = n MF
n= 228/106= 2.15~~2
MF = Na2CO3.5
Similar questions