\(-\frac{25}{35}\cdot\frac{5}{6}< x< (-2)^2\)

\(\Rightarrow-\frac{25}{7}\cdot\frac{1}{6}< x< 4\)

\(\Rightarrow-\frac{25}{42}< x< 4\)

Tự tìm x nhé :v nếu x không có thì ghi : x không thỏa mãn điều kiện

Đổi : \(40(ml)=0,04(l) , 200(ml)=0,2(l) \)

\(\Rightarrow n_{H_2SO_4} (bđ)=0,04.3=0,12(mol) \)

\(\Rightarrow CM_{H_2SO_4}=\dfrac{0,12}{0,2}=0,6(M)\)

\(33,5.3+33.5.x=335\)

\(33,5.\left(3+x\right)=335\)

\(3+x=10\)

\(x=7\)

x = 7

t mik nha

35.5 x 3 +35.5 x X=335

175x3+175x X=335

175x(3+x)=335

3+x=335:175

3+x=\(\frac{67}{35}\)

x=\(\frac{67}{35}\) -3

x=\(\frac{-38}{35}\)

a) Ta có: \(4\cdot\left(13-16\right)-\left(3-5\right)\cdot\left(-3\right)^2\)

\(=4\cdot13-4\cdot16+2\cdot9\)

\(=52-64+18\)

=6

b) Ta có: \(-\dfrac{2}{5}+\dfrac{7}{11}+\dfrac{-11}{10}+\dfrac{7}{-11}\)

\(=-\left(\dfrac{2}{5}+\dfrac{11}{10}\right)+\left(\dfrac{7}{11}-\dfrac{7}{11}\right)\)

\(=-\left(\dfrac{4}{10}+\dfrac{11}{10}\right)\)

\(=\dfrac{-15}{10}=\dfrac{-3}{2}\)

\(PTK_X=102\left(đvC\right)\\ \Rightarrow M_X=102\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow m_O=\%O.M_X=47,06\%.102=48\left(g\right)\)

\(\Rightarrow n_O=\dfrac{m}{M}=\dfrac{48}{16}=3\left(mol\right)\)

\(\Rightarrow CTHH.của.M.có.dạng:X_2O_3\)

\(\Leftrightarrow X.2+16.3=102\\ \Leftrightarrow X=27\left(đvC\right)\)

\(\Rightarrow X.là.Al\left(nhôm\right)\)

\(\Rightarrow CTHH.của.M:Al_2O_3\)

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{4.8}{24}=0,2mol\) 

đổi 200 ml = 0,2 l

PTHH: Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

TL;        1        2           1          1

mol:     0,2 \(\rightarrow\) 0,2               \(\rightarrow\) 0,2

b. \(C_{M_{ddHCl}}=\dfrac{n_{HCl}}{V_{dd_{HCl}}}=\dfrac{0,2}{0,2}=1M\)  

\(c.V_{H_2}=n_{H_2}.22,4=0,2.22,4=4,48l\) 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a) PTHH : \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)

c) \(n_{Mg}=n_{H2}=0,2\left(mol\right)\Rightarrow V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

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Name of school( tên trường của bạn) : VD: junior hight Song Da 
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Reasons you don`t like it ( lí do trường bạn không thích) : VD: Because l have a lot of schools and students are busy taking beautiful car in the garage.

a. \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b. \(m_{HCl}=\dfrac{100\cdot9,6}{100}=9,6\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{9,6}{36,5}\approx0,3\left(mol\right)\)

Theo PTHH: \(n_{Mg}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0,3=0,15\left(mol\right)\)

\(\Rightarrow a=0,15\cdot24=3,6\left(g\right)\)

c.Theo PTHH: \(n_{H2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0,3=0,15\left(mol\right)\)

\(\Rightarrow V_{H2}=0,15\cdot22,4=3,36\left(l\right)\)

a) PTHH: Mg + 2HCl ->MgCl2 + H2

b) nHCl= 100:36,5=2,74(mol)

pthh: Mg + 2HCl -> MgCl2 + H2

-> nMg= 2,74÷2×1=1,37(mol)

-> mMg=1,37×24=32,88(g)

-> a=32,88(g)

c) Theo ý a) -> nH2=2,74(mol)

->VH2=2,74×22,4=61,376(l)

a) Mg + 2HCl -> MgCl2 + H2

b) Theo đầu bài ->mHCl=9,6(g)

-> nHCl= 9,6÷36,5=0,3(mol)

Theo PTHH ta có nMg=nHCl×1/2= 0,3×1/2=0,15(mol)

-> a=0,15×24=3,6(g)

c) Theo PTHH ta có: nH2= nHCl×1/2=0,3×1/2=0,15(mol)

-> VH2= 0,15×22,4=3,36(l)