Question

Answer 1

Lời giải:

\(n_{Glucozo}=n_{Br_2}=0,005\left(mol\right)\\ n_{Ag}=2n_{Glucozo}+2n_{Frucozo}=0,04\left(mol\right)\\ \Rightarrow n_{Frucozo}=0,015\left(mol\right)\\ \Rightarrow Glucozo=25\%\text{ }\text{ }\text{ }\text{ },Frucozo=75\%\)

Answer 2

$n_{glucozo} = n_{Br_2} = \dfrac{0,8}{160} = 0,005(mol)$

$n_{Ag} = \dfrac{4,32}{108} = 0,04(mol)$
$2n_{glucozo} + 2n_{fructozo} = n_{Ag}$

$\Rightarrow n_{frutozo} = \dfrac{1}{2}(0,04 - 0,005.2 )= 0,015(mol)$

Vậy  :

$\%m_{fructozo} = \dfrac{0,015}{0,015 + 0,005} = 75\%$

Đáp án C

Answer 3

\(C.75\%\)