Ta có: 392n_Cr2(SO4)3 + 152n_FeSO4 = 67,2 (1)

PT: \(2Cr+3H_2SO_4\rightarrow Cr_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=3n_{Cr_2\left(SO_4\right)_3}+n_{FeSO_4}=0,32.1,5=0,48\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Cr_2\left(SO_4\right)_3}=0,09\left(mol\right)\\n_{FeSO_4}=0,21\left(mol\right)\end{matrix}\right.\)

Theo PT: n_Cr = 2n_Cr2(SO4)3 = 0,18 (mol)

n_Fe = n_FeSO4 = 0,21 (mol)

⇒ n_Cr:n_Fe = 0,18:0,21 = 6:7

2-4-6

chắc vậy

\(1\) \(-\) \(3\) \(-\) \(5\)

      \(=\)\(>\) \(2\) \(-\) \(4\) \(-\) \(6\)

chắc quan hệ là từ đòng âm

lễ phép ,khiêm nhường ,tôn trọng, tinh tế ,tài hoa, khé léo

(đúng cho mình xin 1 tick nha)

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vô đó mà xem

x+2/5=7/10

x=7/10-2/5

x=7/20-4/10

x=3/10

vậy x=3/10

3/10

Ta có: \(11:\left(x-9\right)=350\)

=>\(x-9=\dfrac{11}{350}\)

=>\(x=\dfrac{11}{350}+9=\dfrac{3161}{350}\)

Ta có: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)

=>\(\left[{}\begin{matrix}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{7}{5}x-\dfrac{4}{3}x=-\dfrac{1}{4}-\dfrac{2}{3}\\\dfrac{7}{5}x+\dfrac{4}{3}x=\dfrac{1}{4}-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{15}x=\dfrac{-11}{12}\\\dfrac{41}{15}x=\dfrac{-5}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{12}:\dfrac{1}{15}=-\dfrac{11}{12}\cdot15=-11\cdot\dfrac{5}{4}=-\dfrac{55}{4}\\x=-\dfrac{5}{12}:\dfrac{41}{15}=-\dfrac{5}{12}\cdot\dfrac{15}{41}=\dfrac{-5\cdot5}{4\cdot41}=\dfrac{-25}{164}\end{matrix}\right.\)

5 lần số lớn là 98-5,5=92,5

Số lớn là 92,5:5=18,5

Số bé là 18,5-5,5=13