Question

ME AJUDEM!!
Determine a fórmula centesimal das seguintes substâncias:

a) H3PO4
b) Ca(OH)2
c) MgSO4
d) H2SO4
e) KMnO4
f) FeCl2


Dadas as massas atômicas: H = 1, P = 31, O = 16, Ca = 40, Mg = 24, S = 32, K = 39, Mn = 55, Fe = 56 Cl = 35,5.

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Answer 1

Resposta:Segue as contas abaixo na explicação

Explicação:

a)H3PO4--->1.3+31.1+16.4-->3+31+64-->98 g/mol

T1=m1/ms/1/100      T2=m2/ms/1/100      T3=m3/ms/1/100

T1=3/98/1/100         T2=31/98/1/100        T3=64/98/1/100

T1=300/98               T2=3100/98              T3=6400/98

T1=3%                       T2=32%                     T3=65%

b)Ca(OH)2-->40.1+16.2+1.2-->40+32+2-->74 g/mol

T1=m1/ms/1/100      T2=m2/ms/1/100      T3=m3/ms/1/100

T1=40/74/1/100       T2=32/74/1/100        T3=2/74/100

T1=4000/74             T2=3200/74              T3=200/74

T1=54%                     T2=43%                     T3=3%

c)MsSO4--->24.1+32.1+16.4-->24+32+64-->120 g/mol

T1=m1/ms/1/100      T2=m2/ms/1/100      T3=m3/ms/1/100

T1=24/120/1/100     T2=32/120/1/100      T3=64/120/1/100

T1=2400/120           T2=3200/120            T3=6400/120

T1=20%                     T2=27%                     T3=53%

d)H2SO4--->1.2+32.1+16.4-->2+32+64-->98 g/mol

T1=m1/ms/1/100      T2=m2/ms/1/100      T3=m3/ms/1/100

T1=2/98/1/100         T2=32/98/1/100        T3=64/98/1/100

T1=200/98               T2=3200/98              T3=6400/98

T1=2%                       T2=33%                      T3=65%

e)KMnO4-->39.1+55.1+16.4-->39+55+64--->158 g/mol

T1=m1/ms/1/100      T2=m2/ms/1/100      T3=m3/ms/1/100

T1=39/158/1/100     T2=55/158/1/100      T3=64/158/1/100

T1=3900/158           T2=5500/158            T3=6400/158

T1=24,70%               T2=34,80%                T3=40,50%

f)FeCl2-->56.1+35,5.2--->56+71--->127 g/mol

T1=m1/ms/1/100        T2=m2/ms/1/100

T1=56/127/1/100       T2=71/127/1/100

T1=5600/127             T2=7100/127

T1=44%                      T2=56%