Sao lại có chuyện tan hết rồi tăng lại được em.

Xuất hiện kết tủa keo trắng

\(AlCl_3+3KOH\rightarrow Al\left(OH\right)_3+3KCl\)

1. businessman

2. musician

3. millionaire

Em tham khảo !

\(\%^{25}Mg=x\left(\%\right)\)

\(\%^{26}Mg=y\left(\%\right)\)

\(TC:\)

\(79\%+x+y=100\%\)

\(\Leftrightarrow x+y=0.21\left(1\right)\)

\(\overline{A_{Mg}}=\dfrac{79\%\cdot24+25\cdot x\%+26\cdot y\%}{100\%}=24.305\)

\(\Rightarrow25x+26y=5.345\left(2\right)\)

\(\left(1\right),\left(2\right):x=0.115,y=0.095\)

\(\%^{25}Mg=11.5\left(\%\right)\)

\(\%^{26}Mg=9.5\left(\%\right)\)

study

a. 

K(SO₄) => K₂SO₄

b.

CuO₃ => Cu₂O hoặc CuO 

d. 

Ag₂NO₃ => AgNO₃

\(n_{BaCl_2}=\dfrac{20.8}{208}=0.1\left(mol\right)\)

\(n_{MgSO_4}=\dfrac{18}{120}=0.15\left(mol\right)\)

\(BaCl_2+MgSO_4\rightarrow BaSO_4+MgCl_2\)

\(0.1..............0.1.............................0.1\)

\(n_{MgSO_4\left(dư\right)}=0.15-0.1=0.05\left(mol\right)\)

\(MgCl_2\rightarrow Mg^{2+}+2Cl^-\)

\(0.1...............0.1.........0.2\)

\(MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}\)

\(0.05...........0.05..........0.05\)

\(\left[Mg^{2+}\right]=\dfrac{0.1+0.05}{0.1}=1.5\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.1}{0.1}=1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05}{0.1}=0.5\left(M\right)\)