Oblicz stosunek masowy pierwiastków w:
a) H2O
b) CO2
c) Al2O3
Proooooszę o pomoc. :'c
1)mh= 1 umo=16 umh2o= 1*2+16=182/16=1/81:82)mCO2=12u+2*16u=44umC : mO = 12 : 32mC : mO = 3 : 8
3)2*27u+ 3*16u= 54u+ 48u= 102u 54/ 48= 9:8
a. mH= 2*1=2u
mO=16u
st masowy= 2/16 = 1/8
b. mC=12u
mO2= 2*16u=32u
st masowy= 12/32= 3/8
c.mAl2= 27*2=54u
mO3=16*3=48u
st masowy= 54/48= 9/8
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1)mh= 1 u
mo=16 u
mh2o= 1*2+16=18
2/16=1/8
1:8
2)mCO2=12u+2*16u=44u
mC : mO = 12 : 32
mC : mO = 3 : 8
3)2*27u+ 3*16u= 54u+ 48u= 102u
54/ 48= 9:8
a. mH= 2*1=2u
mO=16u
st masowy= 2/16 = 1/8
b. mC=12u
mO2= 2*16u=32u
st masowy= 12/32= 3/8
c.mAl2= 27*2=54u
mO3=16*3=48u
st masowy= 54/48= 9/8