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Gerbel
@Gerbel
July 2022
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Химия ОРГАНИКА
Задача на карбоновые кислоты
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хихимик
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R-CH(OH)COOH + Na ----> R-CH(ONa)COONa + H2............n(H2) = 0,224/22,4 = 0,01 моль....................n(R-CH(ONa)COONa) =n(H2) = 0,01 моль....................M(R-CH(ONa)COONa) = 1,2/0,01= 120 г/моль........................x+12*2 +16*3 +1+23*2= 120 ........................x=1.....................R = H...................CH2(ONa)COONa + H2O----> CH2(OH)COONa + NaOHn(CH2(ONa)COONa) =n(NaOH)= 0,01 моль....................c(CH2(ONa)COONa) =c(NaOH)= 0,01/0,25= 0,04 моль/л....................
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Answers & Comments
Verified answer
R-CH(OH)COOH + Na ----> R-CH(ONa)COONa + H2............n(H2) = 0,224/22,4 = 0,01 моль....................n(R-CH(ONa)COONa) =n(H2) = 0,01 моль....................M(R-CH(ONa)COONa) = 1,2/0,01= 120 г/моль........................x+12*2 +16*3 +1+23*2= 120 ........................x=1.....................R = H...................CH2(ONa)COONa + H2O----> CH2(OH)COONa + NaOHn(CH2(ONa)COONa) =n(NaOH)= 0,01 моль....................c(CH2(ONa)COONa) =c(NaOH)= 0,01/0,25= 0,04 моль/л....................