|
Name |
Age |
Occupation |
Interests and hobbies |
Additional information |
|
Rupert |
14 |
student |
Skateboarding, cycling, loud music, parties |
Sociable person |
|
Alfred |
Professor of history |
Working on his study on historical documents and ancient manuscripts. Enjoys talking to his grandchildren |
can not stand noise |
|
|
Ken |
student |
Archcology |
Preparing for his exams |
|
|
Michael |
Rock guitarist |
Loves to play the guitar till late at night |
||
|
28 |
Learning English, traveling, singing |
From China, |
||
|
John |
36 |
Teacher of English literature |
Reading, different cultures and languages |
Поделитесь своими знаниями, ответьте на вопрос:
mпр (С6H5NH2)=116,5 г 1. определим теоретическую массу анилина.
W(С6H5NH2)=80%=0,8 W(С6H5NH2)=mпр (С6H5NH2) : mтеор(С6H5NH2)
m(C6H5NO2)-? mтеор(С6H5NH2)=mпр (С6H5NH2) :
W(С6H5NH2)=116,5г:0,8=145,6г
n(С6H5NH2)= mтеор(С6H5NH2):М(С6H5NH2)=145,6г:93г/моль=1,6 моль
М(С6H5NH2)=12*6+5+14+2=93г/моль
3 C6H5NO2 + 3H2 → С6H5NH2 +2H2O
по уравнению: 1 моль(C6H5NO2):1 моль (С6H5NH2)
по условию: х:1,6 моль
1 моль:х=1 моль:1,6 моль
х=1,6 моль. n(С6H5NO2)=1,6 моль
n(С6H5NO2)=m(C6H5NO2):М(С6H5NО2)
М(С6H5NО2)=12*6+5+14+16*2=123г/моль
m(C6H5NO2)=n(С6H5NO2)*М(С6H5NО2)=1,6 * 123 = 196,8г
Ответ: m(C6H5NO2)=196,8г.