Oblicz masy cząsteczkowe i procentowe zawartości węgla w związkach CH4,C2H6,C2H4,C2H2. POMOCY!!!!!!!
CH4
mC= 12u
mH = 4u
mCH4 = 12u + 4u = 16u
C% = 12/16 * 100% = 75%
C2H6
mC = 2 * 12u = 24u
mH = 6u
mC2H6 = 24u + 6u = 30u
C% = 24/30 *100% = 80%
C2H4
mC2H6 = 24u + 4u = 28u
C% = 24/28 * 100% = 85,71%
C2H2
mC = 2 * 12 = 24u
mH = 2u
mC2H2 = 24u + 2u = 26u
C% = 24/26 * 100% = 92,31%
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CH4
mC= 12u
mH = 4u
mCH4 = 12u + 4u = 16u
C% = 12/16 * 100% = 75%
C2H6
mC = 2 * 12u = 24u
mH = 6u
mC2H6 = 24u + 6u = 30u
C% = 24/30 *100% = 80%
C2H4
mC = 2 * 12u = 24u
mH = 4u
mC2H6 = 24u + 4u = 28u
C% = 24/28 * 100% = 85,71%
C2H2
mC = 2 * 12 = 24u
mH = 2u
mC2H2 = 24u + 2u = 26u
C% = 24/26 * 100% = 92,31%