Question

thành phần % m_H trong hợp chất C_nH_2n+1OH là 13,51%
tìm n

Answer 1

Answer 2

Theo đề bài ta có:

\(\frac{2n+1+1}{12n+2n+1+17}.100\%=13,51\left(\%\right)\)

\(\Leftrightarrow\frac{2\left(n+1\right)}{2\left(7n+9\right)}=\frac{13,51}{100}\\ \Leftrightarrow\frac{n+1}{7n+9}=\frac{13,51}{100}\\ \Leftrightarrow100n+100=94,57n+121,59\\ \Leftrightarrow5,43n=21,59\\ \Leftrightarrow n=\frac{21,59}{5,43}\approx4\)

Answer 3

%H = (2n+2)/(14n+18) *100% = 13.51%

=> n = 4