Determinati formula procentuala, raportul atomic, si raportul de masa pentru urmatoarele substante:
Bioxid de carbon, hidroxid de calciu, pentaoxid de azot, trioxid de azot, apa, apa oxigenata (H2O2)
Răspunsuri la întrebare
Răspuns: CO2 MCO2=12+32=44g/mol C:O=1:2 raport atomic C:O 12:32=3:8 simplificare cu 4 44g......12gC........32gO 100g.....x...........y x=27,27%C y=72,73%O MCa[OH}2= 40+32+2=74g/mol Ca:O:H=1:2:2 Ca:o:H=40:32:2=20:16:1 simplificcu2 74G...........40gCa ........32gO...........2gH 100...........x................y..................z x=54,05%Ca y=43,24%O z=2,75% N2O5 M=14.2+5.16=108g/mol N:O=2:5 N:O=2.14:5.16=28:80=7:20 108g... 28gN..............80gO ..100..........x......................y x=25,92%N y=74,08%O N2O3 M=14.2=3.16=76g/mol N:O=2:3 N:O=28:48=7:12 simplific cu 4 76g...............28gN............48gO 100g............x................y x=36,84%N y=63,16%O H2O MH2O=18g/mol H;O=1:2 H:O=2:16=1:8 simplific cu 2 18g.......2gH.......16gO 100g.....x..........y x=11,11%H y=88,89%O H2O2 M=2+32=34G/mol H:O=2:2=1:1 H:O=2:32=1:16 34g......2gH...........32gO 100g........x..............y x=5,88%H y=94,12%O
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