HOC24
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\(n_{Fe}=\frac{a}{56}\left(mol\right);n_{Al}=\frac{b}{27}\left(mol\right)\)
\(PT:Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(2\right)\)
\(n_{H_2\left(1\right)}=n_{Fe}=\frac{a}{56}\left(mol\right)\)
\(n_{H_2\left(2\right)}=\frac{3}{2}n_{Al}=\frac{b}{18}\left(mol\right)\)
Vì 2 cốc đựng dung dịch HCl và dung dịch H2SO4 ở vị trí thăng bằng, mà cho a gam Fe vào cốc đựng dung dịch HCl, cho b gam Al vào cốc đựng dung dịch H2SO4, cân vẫn ở vị trí thăng bằng nên \(m_{Fe}-m_{H_2\left(1\right)}=m_{Al}-m_{H_2\left(2\right)}\)
\(\Leftrightarrow a-\frac{a}{56}.2=b-\frac{b}{18}.2\)
\(\Leftrightarrow a-\frac{a}{28}=b-\frac{b}{9}\)
\(\Leftrightarrow\frac{27}{28}a=\frac{8}{9}b\)
\(\Rightarrow\frac{a}{b}=\frac{\frac{8}{9}}{\frac{27}{28}}=\frac{224}{243}\)
c) \(PT:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,2 0,2 (mol)
\(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)
b) \(n_{H_2SO_4\left(\frac{1}{2}ddA\right)}=\frac{0,4}{2}=0,2\left(mol\right)\)
\(PT:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
a) \(n_{SO_3}=\frac{32}{80}=0,4\left(mol\right);n_{H_2O}=\frac{68}{18}=\frac{34}{9}\left(mol\right)\)
\(\Rightarrow SO_3\) P/ư hết, \(H_2O\) dư
\(PT:SO_3+H_2O\rightarrow H_2SO_4\)
0,4 0,4 (mol)
\(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
\(m_{dd}=32+68=100\left(g\right)\)
\(\Rightarrow C\%_{ddH_2SO_4}=\frac{39,2}{100}.100\%=39,2\%\)
\(m_{ddH_2SO_4}=10\left(t\right)=10000000\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}=\frac{10000000.98\%}{100\%}=9800000\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{9800000}{98}=100000\left(mol\right)\)
\(PT:Ca_3\left(PO_4\right)_2+2H_2SO_4\rightarrow Ca\left(HPO_4\right)_2+2CaSO_4\)
100000 50000 100000 (mol)
\(m_{Ca\left(HPO_4\right)_2}=50000.232=11600000\left(g\right)=11,6\left(t\right)\)
\(m_{CaSO_4}=100000.136=13600000\left(g\right)=13,6\left(t\right)\)
\(\Rightarrow m_{superphotphatđơn}=11,6+13,6=25,2\left(t\right)\)
\(\Rightarrow H_{pư}=\frac{20,3}{25,2}.100\%=80,56\%\)
\(\text{Đkxđ:}\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
\(A=\frac{\sqrt{a}-2}{1-\sqrt{a}}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)
\(=\frac{2-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{-\left(a-4\right)-\left(a-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{-a+4-a+1+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
Đặt A=\(\sqrt{3+\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}\)
\(=\sqrt{5}+1\)
\(\Rightarrow A=\frac{\sqrt{5}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{10}+\sqrt{2}}{2}\)
Đặt A=\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\Leftrightarrow\frac{1}{\sqrt{2}}A=\frac{1}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{1}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{1}{2+\sqrt{4+2\sqrt{3}}}+\frac{1}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{1}{2+\sqrt{3+2\sqrt{3}+1}}+\frac{1}{2-\sqrt{3-2\sqrt{3}+1}}\)
\(=\frac{1}{2+\sqrt{3}+1}+\frac{1}{2-\sqrt{3}+1}\)
\(=\frac{1}{3+\sqrt{3}}+\frac{1}{3-\sqrt{3}}\)
\(=\frac{3-\sqrt{3}+3+\sqrt{3}}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6}{9-3}\)
= 1
\(\Rightarrow A=\sqrt{2}\)
\(=\sqrt{5}\left(\sqrt{6}+1\right).\frac{\sqrt{2\sqrt{3}-\sqrt{2}}}{\sqrt{2\sqrt{3}+\sqrt{2}}}\text{}\text{}\text{}\text{}\text{}\)
\(=\sqrt{5}\left(\sqrt{6}+1\right).\frac{\sqrt{\sqrt{2}\left(\sqrt{6}-1\right)}}{\sqrt{\sqrt{2}\left(\sqrt{6}+1\right)}}\)
\(=\frac{\sqrt{5}\left(\sqrt{\sqrt{6}+1}\right)^2.\sqrt{\sqrt{2}}.\sqrt{\sqrt{6}-1}}{\sqrt{\sqrt{2}}.\sqrt{\sqrt{6}+1}}\)
\(=\sqrt{5}\left(\sqrt{\sqrt{6}+1}\right)\sqrt{\sqrt{6}-1}\)
\(=\sqrt{5}.\sqrt{6-1}\)
=\(\sqrt{5}.\sqrt{5}\)
\(=5\)
\(\sqrt{5}\left(\sqrt{6}+1\right):\frac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}-\sqrt{2}}}\)