Prosze o pomoc. Oblicz procent masowy magnezu w następujących solach :
a)Mg(No3)2
b)MgS.
Dam Naj!!
a)
Mg(NO3)2 = 24u+ 2*14u + 6*16u = 148u
148u -> 100%
24u -> x
x=16,22% Mg
b)
MgS= 24u + 32u = 56u
56u -> 100%
24u-> x
x=42,86% Mg
mMg(NO3)2=24u+2*14u+6*16u=148u
148u-------100%
24u--------x
x=16,2% stanowi Mg w Mg(NO3)2
mMgS=24u+32u=56u
56u------100%
24u------x
x=42,9% stanowi Mg w MgS
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a)
Mg(NO3)2 = 24u+ 2*14u + 6*16u = 148u
148u -> 100%
24u -> x
x=16,22% Mg
b)
MgS= 24u + 32u = 56u
56u -> 100%
24u-> x
x=42,86% Mg
mMg(NO3)2=24u+2*14u+6*16u=148u
148u-------100%
24u--------x
x=16,2% stanowi Mg w Mg(NO3)2
mMgS=24u+32u=56u
56u------100%
24u------x
x=42,9% stanowi Mg w MgS