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alina28042002
@alina28042002
August 2022
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вычислите массовую долю элемента кислород: а) в воде; б) в кварце SiO2; в) в серной кислотеH2SO4; г) в соде Na2CO3.
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Alexei78
Verified answer
A) H2O
M(H2O)=18 g/mol
Ar(O)=16
W(O)=Ar / Mr * 100% = 16 / 18*100%=88.89%
б) SiO2
M(SiO2)=28*16*2=60 g/mol
W(O) = 16*2 / 60*100%=53.33 %
в) H2SO4
M(H2SO4)=98 g/mol
W(O)=16*4 / 98*100%=65.3%
г) Na2CO3
M(Na2CO3)=23*2+12+16*3=106 g/mol
W(O)=16*3 / 106*100%= 45.3%
1 votes
Thanks 1
JeTime
Verified answer
A)Mr ( H2O) = (1*2) + 16 = 18
W (O) = 16/18 * 100% = 88.8 ~ 89%
б)Mr (SiO2) = 28 + (16*2) = 60
W(O) = 32/60 * 100% = 53.3%
в)Mr ( H2SO4) = (1*2) + 32 + (16*4) = 98
W(O) = 64 / 98 * 100% = 65.3%
г)Mr (Na2CO3) = (23 * 2) + 12 + (16 * 3) = 106
W (O) = 48 / 106 * 100% = 45.2%
1 votes
Thanks 0
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Answers & Comments
Verified answer
A) H2OM(H2O)=18 g/mol
Ar(O)=16
W(O)=Ar / Mr * 100% = 16 / 18*100%=88.89%
б) SiO2
M(SiO2)=28*16*2=60 g/mol
W(O) = 16*2 / 60*100%=53.33 %
в) H2SO4
M(H2SO4)=98 g/mol
W(O)=16*4 / 98*100%=65.3%
г) Na2CO3
M(Na2CO3)=23*2+12+16*3=106 g/mol
W(O)=16*3 / 106*100%= 45.3%
Verified answer
A)Mr ( H2O) = (1*2) + 16 = 18W (O) = 16/18 * 100% = 88.8 ~ 89%
б)Mr (SiO2) = 28 + (16*2) = 60
W(O) = 32/60 * 100% = 53.3%
в)Mr ( H2SO4) = (1*2) + 32 + (16*4) = 98
W(O) = 64 / 98 * 100% = 65.3%
г)Mr (Na2CO3) = (23 * 2) + 12 + (16 * 3) = 106
W (O) = 48 / 106 * 100% = 45.2%