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Lafolledu974
@Lafolledu974
April 2019
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Physique . Exercice sur les masse molaire urgent svp
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Kilaro
Exercice 1 )
m=0.20g
M(Au) = 197g.mol-1
On sait que n=m/M; donc on applique
n=1.0*10^-3 mol
Exercice 2)
2.1)M(C6H12O6) = M(C)*6 + M(H)*12 + M(O)*6 soit
M(C6H12O6)=72+12+96 = 180g.mol-1
2.2)meme formule que l'exo 1!
n=2.50/180=1.4*10^-2 mol
Exercice 3
3.1)encore la formule n=m/M donc :
n=20.0/63.5=3.14*10^-1 mol
3.2)M(CO2) = M(C) + M(O)*2
M(CO2)=44g.mol-1
3.3)encore n=m/M :
n=30.0/44=6.8*10^-1mol
Exo 4 apres
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Kilaro
j'avais pu de temps ! donc exercice 4 :
Kilaro
4.1)M(H3BO3)=M(H)*3+M(B)+M(O)*3 = 62g.mol-1
Kilaro
4.2)ici la formule c'est C=n/V donc n=C*V: n=2.6*10^-2*2.5*10^-1 = 6.5*10^-3 mol
Kilaro
4.3)on connais n on connais M donc n=m/M soit m=M*n;on obtient :
Kilaro
a la 4.1) c'est 63
Kilaro
non c'est bien 62 mais jenetrouve pas pour la derniere enfin je trouve 1.69*10^-4g bizarre non ?
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Lista de comentários
m=0.20g
M(Au) = 197g.mol-1
On sait que n=m/M; donc on applique
n=1.0*10^-3 mol
Exercice 2)
2.1)M(C6H12O6) = M(C)*6 + M(H)*12 + M(O)*6 soit
M(C6H12O6)=72+12+96 = 180g.mol-1
2.2)meme formule que l'exo 1!
n=2.50/180=1.4*10^-2 mol
Exercice 3
3.1)encore la formule n=m/M donc :
n=20.0/63.5=3.14*10^-1 mol
3.2)M(CO2) = M(C) + M(O)*2
M(CO2)=44g.mol-1
3.3)encore n=m/M :
n=30.0/44=6.8*10^-1mol
Exo 4 apres