Предмет: Химия, автор: ns069671

Густина парів речовини за воднем становить 45. масові частки карбону гідрогену й оксигену у сполуці відповідно становлять 0,2667 , 0,0224 і 0,7109 . установіть молекулярну формулу сполуки

Ответы

Автор ответа: unezhewaoxnw6u

Ответ:

Объяснение:

CxHyOz=x : y : z= 0,2667/12 : 0,0224/1 : 0,7109/16 *100% =0,0222 : 0,0224 :

0,0444 = 1 : 1 : 2

М(СНО2) = 45 г/моль

М(CxHyOz)=45*2=90г/моль

90:45=2

ОТВЕТ:С2Н2О4 'то щавелевая кислота, HOOC−COOH

Ar(C) = 12 г/моль

Ar(H) = 1 г/моль

Ar(O) = 16 г/моль

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