1.Calculeaza:
a)numarul de molecule din 1,5 moli amoniac
b)compozitia procentuala a H2S

2.Calculeaza raportul numarului de atomi,respectiv,raportul masic,pentru urmatoarele substante compuse:
a)SiCl4
b)KClO3

3.In ce masa de hidroxid de calciu se gasesc:
a)90g calciu
b)25g oxigen

2

Răspunsuri

2014-05-01T18:11:26+03:00
1)a)1 mol NH3........6,022 * 10 la 23 molec
      1,5 moli NH3............x molec        x = 9,033 * 10 la 23 molec
b)masa molara H2S =2*A hidrogen + A sulf=2+32=34 g/mol
%H=masa hidrogen/masa molara H2S*100=2/34*100=5,88%
%S=masa sulf/masa molara H2S*100=32/34*100=94,118%

2)a)raport atomic 1:4 si raportul masic 28:142=14:71
b)raport atomic 1:1:3 si raportul masic 39:35,5:48

3)masa molara Ca(OH)2=A calciu + 2* A oxigen + 2 * A hidrogen = 40+2+32=74 g/mol
a)74 g Ca(OH)2.........40 g Ca
   a g Ca(OH)2............90 g Ca      a = 166,5 g Ca
b)74 g Ca(OH)2.........32 g O
   b g Ca(OH)2...........25 g O      b = 57,8125 g O
2 5 2
Cel mai inteligent răspuns!
2014-05-01T18:13:04+03:00
1.a)NH3-amoniac
1mol.................6.023·10²³molecule
1.5moli..................x
x=1.5·6.023·10²³/1=9.0345·10²³
b)Masa molara H2S=2*1+1*32=34g
34gH2S..................2gH..............32gS
100gH2S..................x.................y
x=100*2/34=5.88%H
y=100*32/34=94.11%S

2.a)SiCl4
Si:Cl=1:4
Si:Cl=28:35.5
b)KClO3
K:Cl:O=1:1:3
K:Cl:O=39:35.5:16

3.Masa molara Ca(OH)2=1*40+2*16+2*1=73g
a)73gCa(OH)2...............40gCa
x.................................90gCa
x=73*90/40=164.25gCa(OH)2
b)73gCa(OH)2....................32gO
x.......................................25gO
x=73*25/32=57.03gCa(OH)2
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