Oblicz masy molowe następujących substancji:
a) NH3 b)Fe(SO4)3 c)Ba(OH)2 d)CO2 e)SO3 f)(NH4)2SO4
g)MgCO3 h)K3PO4
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Masa N = 14g
Masa H = 1g
Masa C = 12g
Masa S = 32g
Masa O = 16g
Masa Ba = 137,33g
Masa K = 39g
Masa P = 30,97g
Masa Fe = 55,85g
Masa Mg = 24,3g
a) NH3 = 14g+3*1g = 17g
b)Fe2(SO4)3 = 2*55,85g + 3*32g+12*16g = 399,7g
c)Ba(OH)2 = 137,33g+2*16g+2*1g = 171,33g
d)CO2 = 12g+2*16g = 44g
e)SO3 = 32g+3*16g = 80g
f)(NH4)2SO4 = 2*14g+8*1g+32g+4*16g = 132g
g)MgCO3 = 24,3g+12g+3*16g = 84,3g
h)K3PO4 = 3*39g+30,97g+4*16g = 211,97g