Răspunsuri

2014-05-08T21:28:53+03:00
in 63g HNO3...1gH...14gN...48gO2
in 100g HNO3..xgH..ygN..zgO2
facand regula de 3 simpla avem:
63/100=1/x  => x=100/63=1,59 H     (prin aproximare)
luam iar 63/100,dar cu N..deci avem:
63/100=14/y , 63y=1400=> y=1400/63=22,22%N
si cu O2:
63/100=48/z, 63z=4800=> z=4800/63=76,2%



in 106g H3PO4...3gH..39gP..64gO2
in 100g H3PO4..xgH...ygP....zgO2
106/100=3/x, 106x=300 => x=300/106=2,83%H
106/100=39/y, 106y=3900 => y=3900/106=36,8% P
106/100=64/z , 106z=6400 => 6400/106=60,38% O2

daca te intrebi de unde am luat..63g...si 106..atunci  uite:
MmolaraHNO3=63g
MmolaraH3PO4=106g. celelalte le-am scos de la masa atomica rotunjita a elementellor:
de exemplu: 14gN=masa atomica a Azotului :) sper ca ti-am fost de ajutor!







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