2 cặp

a. \(n_{CO_2}=0,2\left(mol\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

\(\Rightarrow n_{MgCl_2}=n_{MgCO_3}=n_{CO_2}=0.2\left(mol\right)\)

\(\Rightarrow m_{MgCO_3}=0,2.84=16,8\left(g\right)\)

\(\%m_{MgCO_3}=\dfrac{16,8.100\%}{24}=70\%\\ \%m_{FeO}=100\%-70\%=30\%\)

b. \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(n_{FeO}=\dfrac{24-16,8}{72}=0,1\left(mol\right)=n_{FeCl_2}\)

\(C_{M_{FeCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\)

\(C_{M_{MgCO_3}}=\dfrac{0,2}{0,7}=\dfrac{2}{7}\left(M\right)\)

Bài 3:

theo đề bài ta có:

\(\left\{{}\begin{matrix}2a-3b=0\\5b-7c=0\\3a-7b+5c=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=42\\b=28\\c=20\end{matrix}\right.\)

để cá có thể thở 

Quảng Ninh

\(\Leftrightarrow\dfrac{\left(x+2\right)x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\) ∀x≠{0;2}

\(\Leftrightarrow x^2+2x-\left(x-2\right)=2\\ \Leftrightarrow x^2+2x-x+2-2=0\\ \Leftrightarrow x^2+x=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

xét điều kiện, ta loại x = 0, nhận x = -1

1.Cùng phương

+\(\overrightarrow{a},\overrightarrow{b}\)

+\(\overrightarrow{x},\overrightarrow{y}\)

+\(\overrightarrow{w},\overrightarrow{z}\)

+\(\overrightarrow{u},\overrightarrow{v}\)

2.Cùng hướng

+\(\overrightarrow{a},\overrightarrow{b}\)

+\(\overrightarrow{x},\overrightarrow{y},\overrightarrow{z}\)

3.Ngược hướng:

+\(\overrightarrow{w},\overrightarrow{z}\)

+\(\overrightarrow{u},\overrightarrow{v}\)

4. Bằng nhau:

+\(\overrightarrow{x},\overrightarrow{y}\)

+\(\overrightarrow{w},\overrightarrow{z}\)

+\(\overrightarrow{u},\overrightarrow{v}\)

5. Đối nhau:

+\(\overrightarrow{u},\overrightarrow{v}\)

+\(\overrightarrow{w},\overrightarrow{z}\)

1/9/1958

1. \(K_2CO_3\Rightarrow PTK:138\)

2.\(Al\left(NO_3\right)_3\Rightarrow PTK:213\)

3.\(FeSO_4\Rightarrow PTK:152\)

4.\(R_2O_n\Rightarrow PTK:2.R+16.n\)