Question

рассчитайте относительную молекулярную массу веществ:К2СО3

Answer 1

1)Mr(HBr) = Ar(H) + Ar(Br) = 1 + 80 = 81 
w(H/HBr) = Ar(H)/ Mr(HBr) x 100%  = 1/81 х 100 = 1,23%
w (Br/HBr) = Ar(Br)/ Mr(HBr) x 100%  = 80/81 х 100 = 98,77%
2) Mr(K2CO3) = 2Ar(K) + Ar(C) + 3Ar(O) = 2x39 + 12 + 3x16 = 138
 w (2K/K2CO3) = 2Ar(K)/ Mr(K2CO3) x 100%  = 78/138 х 100 = 56,52%
w (C/K2CO3) = Ar(C)/ Mr(K2CO3) x 100%  = 12/138 х 100 = 8,69% 
w (3О/ K2CO3) = 3Ar(О)/ Mr(K2CO3) x 100%  = 48/138 х 100= 34,79%3)Mr(Mg(OH)2) = 24 + 16x2 + 1x2 = 58
w (Mg/Mg(OH)2) = 24/ 58 х 100 = 41,38%w (2O/Mg(OH)2) = 32/ 58 х 100 = 55,17%
w (2H/Mg(OH)2) = 2/ 58 х 100 = 3,45%4)Mr(P2O5) = 2x31 + 5x16 = 142
w (2P/P2O5) = 2x31/ 142 х 100 = 43,66%
 w (5O/P2O5) = 5x16/ 142 х 100 = 56,34% 5)Mr(feCl3) = 56 + 3 x 35,5 = 162,5  w (Fe/FeCl3) = 56/ 162,5 х 100 = 34,46%
 w (3Cl/FeCl3) = 3x35,5/ 162,5 х 100 = 65,54% 6)Mr(Cu(NO3)2) = 64 + 2x14 + 6x16 = 188
 w (Cu/Cu(NO3)2) = 64/ 188 х 100 = 34,04%
 w (2N/ Cu(NO3)2) = 2x14/ 188 х 100 = 14,9%
w (6O/Cu(NO3)2) = 96/ 188 х 100 = 51,06%  

Answer 2

M(К2СО3)= 2*39+12+3*16=138