a) \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

b) \(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)

c) \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

d) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

e) \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

f) \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

g) Mình sửa HNO₃ thành H₂SO₄ nhá

\(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3\)

h) \(Mg\left(OH\right)_2+2HNO_3\rightarrow Mg\left(NO_3\right)_2+2H_2O\)

i) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

Al với NaOH quất vô nhau ra pêđê /=/

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\)

Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1mol 0,4mol \(\rightarrow\) 0,1mol

Lập tỉ số: \(n_{Fe}:n_{HCl}=0,1< 0,4\)

\(\Rightarrow\)Fe hết, HCl dư

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) Có

\(n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\)

Pt: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,6mol \(\leftarrow\) 0,3mol

\(V_{NaOH}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

4000 + 500 + 60 + 7 = 4567 

tick mình nha

Đề bị sai rồi bạn

Cần bao nhiêu ml NaOH 1M ?

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\(n_{HCl}=0,1.1,5=0,15\left(mol\right)\)

Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,15mol \(\leftarrow\) 0,15mol

\(V_{NaOH}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

Ừ giúp nhanh đi nà :'> Lần ni mà biết kết quả còn đăng lên coi chừng đó

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\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi x, y lần lượt là số mol của Mg, Al

Pt: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)

x \(\rightarrow\) x

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (2)

y \(\rightarrow\) 3y/2

(1)(2)\(\Rightarrow\left\{{}\begin{matrix}x+1,5y=0,4\\24x+27y=7,8\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%Mg=\dfrac{0,1.24}{7,8}.100=30,77\%\)

\(\%Al=\dfrac{0,2.27}{7,8}.100=69,23\%\)

b) \(n_{MgCl_2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(n_{AlCl_3}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

m_Muối = \(0,1.95+0,2.133,5=36,2\left(g\right)\)

( Không thì giải theo định luật bảo toàn khối lượng ..v.v )

2. a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

Vậy...

b) \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow x+1-x^2-2x-1=0\)

\(\Leftrightarrow-x^2-x=0\)

\(\Leftrightarrow-x\left(x+1\right)=0\)

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

Vậy...

c) \(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

Vì \(x^2+1>0\Rightarrow x=0\)

Vậy...

d) \(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)

Vậy..

e) \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

Vậy...

Lần sau ghi tách ra tí bạn ơi ;v

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1. a) \(5x-20y=5\left(x-4y\right)\)

b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)

c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)

\(=\left(x+1\right)\left(x-5\right)\)

d) \(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=4xy\)

e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)

\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)

\(=2x\left(4x+2\right)\)