\(\left(\dfrac{x}{2}-\dfrac{3}{2}\right)\cdot\left(3x-\dfrac{1}{5}\right)=0\)

\(\Rightarrow\) \(\left(\dfrac{x}{2}-\dfrac{3}{2}\right)\) Hoặc \(\left(3x-\dfrac{1}{5}\right)\) = 0

+) \(\dfrac{x}{2}-\dfrac{3}{2}=0\) 

\(\Rightarrow\) \(\dfrac{x}{2}=0+\dfrac{3}{2}\) 

\(\Rightarrow\) \(\dfrac{x}{2}=\dfrac{3}{2}\) 

\(\Rightarrow\) \(x=3\)

+) \(3x-\dfrac{1}{5}=0\)

\(\Rightarrow\) \(3x=\dfrac{1}{5}\)

\(\Rightarrow\)  \(x=\dfrac{1}{5}\div3\)

\(\Rightarrow\) \(x=\dfrac{1}{5}\cdot\dfrac{1}{3}=\dfrac{1}{15}\)

\(\Rightarrow\) \(x\) \(\in\) {\(3;\dfrac{1}{15}\)}

Vậy chọn B 

Khối lượng của hợp kim đó :

\(8,5:\dfrac{5}{8}=13,6\left(kg\right)\)

\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

pt 2CH3COOH+Mg→(CH3COO)2Mg +H2

n(CH3COO)2Mg =1,42/142=0,1 mol

theo pt nCH3COOH =2n(CH3COO)2Mg =0,2 mol

suy ra CM=0,2 /0,5=0.4 mol/l

theo pt nH2 =n(CH3COO)2Mg =0,1 mol

suy ra VH2 =2,24l

KOH+CH3COOH->CH3COOK+H2O

0,2------0,2

=>VKOH=\(\dfrac{0,2}{0,5}\)=0,4l=400ml

Bài 2: 

Số học sinh giỏi là: 300x40%=120(bạn)

Số học sinh nữ đạt loại giỏi là: 120x2/3=80(bạn)

Bài 1: 

a: Số cần tìm là: 51:3/7=119

b: Số cần tìm là -33:11/12=-36

c: Số cần tìm là 7,2:4/5=9

d: Số cần tìm là 36:24%=150

= 150

Bài 2:

\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,05(mol)\\ \Rightarrow m_{Mg}=24.0,05=1,2(g)\\ \Rightarrow m_{MgO}=9,2-1,2=8(g) b,\%_{Mg}=\dfrac{1,2}{9,2}.100\%=13,04\%\\ \Rightarrow \%_{MgO}=100\%-13,04\%=86,96\%\\ c,n_{MgO}=\dfrac{8}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,5(mol)\\ \Rightarrow \Sigma m_{HCl}=0,5.36,5=18,25(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{18,25}{14,6\%}=125(g)\)

Bài 5 :

 \(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2            1              1

      0,05      0,1                          0,05

        \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

         1              2             1            1

        0,2           0,4

a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1,2\left(g\right)\)

\(m_{MgO}=9,2-1,2=8\left(g\right)\)

b) ⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

    ⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

c) Có : \(m_{MgO}=8\left(g\right)\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

 Chúc bạn học tốt

Bài 3:

\(a,PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)\\ \Rightarrow n_{CO_2}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\\ c,m_{dd_{HCl}}=1,15.300=345(g)\\ n_{NaCl}=0,4(mol);n_{H_2O}=n_{CO_2}=0,2(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\\ m_{H_2O}=18.0,2=3,6(g)\\ m_{CO_2}=44.0,2=8,8(g)\\ m_{dd_{NaCl}}=21,2+345-3,6-8,8=353,8(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{23,4}{353,8}.100\%=6,61\%\)

a, \(u_n=u_1.q^{n-1}\)

\(\Leftrightarrow192=u_1.2^n\)

\(\Leftrightarrow u_1=\dfrac{192}{2^n}\)

\(S_n=\dfrac{u_1\left(1-q^n\right)}{1-q}\)

\(\Leftrightarrow189=\dfrac{\dfrac{192}{2^n}\left(1-2^n\right)}{1-2}\)

\(\Leftrightarrow189=192-\dfrac{192}{2^n}\)

\(\Leftrightarrow\dfrac{192}{2^n}=3\)

\(\Leftrightarrow2^n=2^6\)

\(\Rightarrow n=6\)