\(a,\Rightarrow2n+3\ne0\Rightarrow n\ne-\dfrac{2}{3}\\ b,A\in Z\Rightarrow A=\dfrac{6\left(2n+3\right)-17}{2n+3}=6-\dfrac{17}{2n+3}\in Z\\ \Rightarrow2n+3\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\\ \Rightarrow2n\in\left\{-20;-4;-2;14\right\}\\ \Rightarrow n\in\left\{-10;-2;-1;7\right\}\left(tm\right)\)

Câu 2:

Đề là 13,44 lít đk?

\(PTHH:Fe+2HCl\to FeCl_2+H_2\\ n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)\\ \Rightarrow n_{Fe}=n_{H_2}=0,6(mol)\\ \Rightarrow m_{Fe}=0,6.56=33,6(g)\\ \Rightarrow m_{Cu}=50-33,6=16,4(g)\)

Câu 1:

\(n_{HCl}=0,05.2=0,1(mol)\\ \Rightarrow n_{Cl^-}=0,1(mol)\\ PTHH:\\ Mg(OH)_2+2HCl\to MgCl_2+2H_2O\\ Cu(OH)_2+2HCl\to CuCl_2+2H_2O\\ NaOH+HCl\to NaCl+H_2O\\ \Rightarrow n_{OH^-}=n_{Cl^-}=0,1(mol)\\ \Rightarrow m_{OH^-}=0,1.17=1,7(g)\\ \Rightarrow m_{KL}=m_{\text{muối }Cl^-}-m_{Cl^-}=6,025-0,1.35,5=2,475(g)\\ \Rightarrow m_{hh}=m_{KL}+m_{OH^-}=2,475+1,7=4,175(g)\)

\(a,PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O\\ b,n_{Cu}=\dfrac{6,4}{64}=0,1(mol)\\ \text{Theo PT: }n_{H_2}=n_{Cu}=0,1(mol)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24(l)\\ c,n_{CuO}=\dfrac{40}{80}=0,5(mol)\\ \text{Theo PT: }n_{Cu}=n_{CuO}=0,5(mol)\\ \Rightarrow m_{Cu}=64.0,5=32(g)\)

lập tỉ lệ số mol e

\(A:Fe_2O_3\\ B:FeCl_3\\ C:Fe(NO_3)_3\\ D:Fe_2(SO_4)_3\\ E:Na_2SO_4\\ F:NaOH\)

Từ đó ta có các PTHH tương ứng là:

\((1)Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ (2)Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ (3)FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl\downarrow\\ (4)Fe_2(SO_4)_3+6NaOH\to 2Fe(OH)_3\downarrow+3Na_2SO_4\\ (5)Na_2SO_4+Ba(OH)_2\to BaSO_4\downarrow+2NaOH\\ (6)FeCl_3+3NaOH\to Fe(OH)_3+3NaCl\\ (7)\begin{cases} Fe(NO_3)_3+3NaOH\to Fe(OH)_3\downarrow+3NaNO_3\\ NaOH+FeCl_3\to Fe(OH)_3+3NaCl \end{cases}\)

\(PTHH:Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ n_{Al_2O_3}=\dfrac{10,2}{102}=0,1(mol)\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\\ \text{LTL: }\dfrac{0,1}{1}>\dfrac{0,2}{3}\Rightarrow AL_2O_3\text{ dư}\\ \Rightarrow n_{Al_2O_3(dư)}=0,1-\dfrac{0,2}{3}=\dfrac{1}{30}(mol)\\ \Rightarrow m_{Al_2O_3(dư)}=\dfrac{1}{30}.102=3,4(g)\\ b,n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{1}{15}(mol)\\ \Rightarrow m_{Al_2(SO_4)_3}=\dfrac{1}{15}.342=22,8(g)\)

Đặt \(n^2+2021=k^2\left(k\in N\right)\)

\(\Rightarrow k^2-n^2=2021\\ \Rightarrow\left(k-n\right)\left(k+n\right)=2021\)

Mà \(k,n\in N\)

\(\Rightarrow\left(k-n\right)\left(k+n\right)=2021\cdot1=43\cdot47\)

\(\left\{{}\begin{matrix}k-n=2021\\k+n=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=1011\\n=-1010\left(loại\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}k-n=1\\k+n=2021\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=1011\\n=1010\end{matrix}\right.\left(nhận\right)\)

\(\left\{{}\begin{matrix}k-n=43\\k+n=47\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=45\\n=2\end{matrix}\right.\left(nhận\right)\)

\(\left\{{}\begin{matrix}k-n=47\\k+n=43\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=45\\n=-2\left(loại\right)\end{matrix}\right.\)

Vậy \(n\in\left\{2;1010\right\}\)