Question

Answer 1

\(1.\\ a.C\%=\dfrac{\dfrac{10,6}{286}\cdot106}{10,6+64}=5,266\%\\ C_M=\dfrac{0,0370}{0,064+0,0370\cdot10\cdot\dfrac{18}{1000}}=0,524mol\cdot L^{^{ }-1}\\ b.C\%=\dfrac{\dfrac{40}{80}\cdot98}{400}=12,25\%\\ C_M=\dfrac{\dfrac{40}{80}}{0,36}=1,389mol\cdot L^{^{ }-1}\)

Answer 2

\(2.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+H_2SO_4->FeSO_4+H_2\\ m_{H_2SO_4}=0,1\cdot98=9,8g\\ C\%=\dfrac{9,8}{200}=4,9\%\\ V_{H_2}=0,1\cdot22,4=2,24L\\ C\%_{FeSO_4}=\dfrac{0,1\cdot152}{205,6-0,1\cdot2}=7,4\%\)

Answer 3

\(3.Zn+2HCl->ZnCl_2+H_2\\ n_{HCl}=0,2mol\\ m_{Zn}=65\cdot0,2:2=6,5\left(g\right)\\ V_{H_2}=0,2:2\cdot22,4=2,24L\\ C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)

Answer 4

\(4.\\ a.C\%=\dfrac{400\cdot0,196}{600}=13,067\%\\ b.C\%=\dfrac{400\cdot0,196}{300}=26,133\%\)

Answer 5

giúp t vs mn ơi

 

Answer 6

Em cần bài nào vậy em?