Giả sử có 1 lít dd

`=>` \(\left\{{}\begin{matrix}n_{Na^+}=0,01\left(mol\right)\\n_{Mg^{2+}}=0,015\left(mol\right)\\n_{NO_3^-}=0,02\left(mol\right)\\n_{SO_4^{2-}}=x\left(mol\right)\end{matrix}\right.\)

Bảo toàn điện tích:

\(0,01+0,015.2=0,02+2x\\ \Leftrightarrow x=0,01M\)

Vậy chọn C

\(n_{K^+}=1,12.0,25=0,28\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{K_2SO_4}=a\left(mol\right)\\n_{KNO_3}=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}174a+101b=26,04\\2a+b=0,28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,08\left(mol\right)\\b=0,12\left(mol\right)\end{matrix}\right.\\ \Rightarrow n_{SO_4^{2-}}=0,08\left(mol\right)\\ \Rightarrow C_{M\left(SO_4^{2-}\right)}=\dfrac{0,08}{0,25}=0,32M\)

Vậy chọn B

\(n_{Cl^-}=0,5.1,25=0,625\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{BaCl_2}=a\left(mol\right)\\n_{KCl}=3a\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Cl^-}=2n_{BaCl_2}+n_{KCl}=2a+3a=5a\left(mol\right)\\ \Rightarrow5a=0,625\left(mol\right)\\ \Leftrightarrow a=0,125\left(mol\right)\\ \Rightarrow m_{BaCl_2}+m_{KCl}=0,125.208+0,125.3.74,5=53,9375\left(g\right)\)

Chọn C

Có: \(\left\{{}\begin{matrix}2p_A+2p_B+n_A+n_B=94\\2p_A+2p_B-n_A-n_B=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}p_A+p_B=31\left(1\right)\\n_A+n_B=32\end{matrix}\right.\)

Mà \(2p_B-2p_A=14\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}p_A=12\\p_B=19\end{matrix}\right.\)

=> A là Mg, B là K

\(P_{\left(x\right)}=\left(x-4\right)^{\left(x-5\right)\left(x-6\right)\left(x+6\right)\left(x+5\right)}\\ P_{\left(x\right)}=\left(x-4\right)^{\left(x^2-25\right)\left(x^2-36\right)}\)

Thay `x = 7`

\(P_{\left(x\right)}=\left(7-4\right)^{\left(7^2-25\right)\left(7^2-36\right)}\\ =3^{24.13}\\ =3^{312}\)

`a)`

Gọi \(n_K=a\left(mol\right)\)

PTHH:

                   \(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

                   a----------------->a

              \(2KOH+2Al+2H_2O\rightarrow2KAlO_2+3H_2\uparrow\)

                a--------->a

\(\Rightarrow\dfrac{n_K}{n_{Al}}=\dfrac{a}{a}=\dfrac{1}{1}\)

`b)` 

Gọi \(\left\{{}\begin{matrix}n_K=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow39a+27b=14,4\left(1\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH:

                      \(2K+2HCl\rightarrow2KCl+H_2\uparrow\)

                      a--------------------------->0,5a

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

                      b------------------------------>1,5b

\(\Rightarrow0,5a+1,5b=0,3\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,3\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_K=0,3.39=11,7\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)

`c)`

\(n_{Al\left(OH\right)_3}=\dfrac{1,56}{78}=0,02\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT Al}}n_{NaAlO_2}=0,1-0,02=0,08\left(mol\right)\)

\(n_{HCl\left(bđ\right)}=2.0,4=0,8\left(mol\right)\)

PTHH:

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

                      0,1---->0,3------>0,1

                      \(2K+2HCl\rightarrow2KCl+H_2\)

                      0,3-->0,3-------->0,3

 \(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,3-0,3=0,2\left(mol\right)\)

PTHH: 

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

                        0,2<------0,2

                       \(3NaOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

                        0,3<-------0,1

                       \(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)

                        0,08<--------------------0,08

\(\Rightarrow C_{M\left(NaOH\right)}=\dfrac{0,3+0,2+0,08}{2}=0,29M\)

`a)`

\(\left\{{}\begin{matrix}m_{AgNO_3}=\dfrac{34}{100}.100=34\left(g\right)\\m_{BaCl_2}=\dfrac{20,8}{100}.0=10,4\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{AgNO_3}=\dfrac{34}{170}=0,2\left(mol\right)\\n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\end{matrix}\right.\)

PTHH:              \(2AgNO_3+BaCl_2\rightarrow2AgCl\downarrow+Ba\left(NO_3\right)_2\)

ban đầu            0,2             0,05 

sau pư               0,1                0           0,1             0,05

`b)`

\(m_{dd}=100+50-0,1.143,5=135,65\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{AgNO_3\left(dư\right)}=\dfrac{0,1.170}{135,65}.100\%=12,53\%\\C\%_{Ba\left(NO_3\right)_2}=\dfrac{0,05.261}{135,65}.100\%=9,62\%\end{matrix}\right.\)

`1)` Có: \(\left\{{}\begin{matrix}p+e+n=49\\n=53,125\%.\left(p+e\right)\\p=e\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}p=e=16\\n=17\end{matrix}\right.\)

`2)` Lưu huỳnh - `S`

`a)`

\(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{a}{100}\left(mol\right)\\n_{M_2CO_3}=\dfrac{b}{2M_M+60}\left(mol\right)\end{matrix}\right.\xrightarrow[]{\text{BTNT C}}\left\{{}\begin{matrix}n_{CO_2\left(A\right)}=\dfrac{a}{100}\left(mol\right)\\n_{CO_2\left(B\right)}=\dfrac{b}{2M_M+60}\left(mol\right)\end{matrix}\right.\)

Có:

- Lượng H₂SO₄ bằng nhau và dư

- Lúc sau cân lại trở về vị trí cân bằng

`=>` \(m_{CaCO_3}+m_{H_2SO_4}-m_{CO_2\left(A\right)}=m_{M_2CO_3}+m_{H_2SO_4}+m_{CO_2\left(B\right)}\)

`=>` \(a-44.\dfrac{a}{100}=b-44.\dfrac{b}{2M_M+60}\)

`<=>` \(M_M=\dfrac{33,6a-16b}{2b-1,12a}\left(g\text{/}mol\right)\)

`b)`

\(M_M=\dfrac{33,6.5-16.4,8}{2.4,8-1,12.5}=22,8\approx23\left(g\text{/}mol\right)\)

Vậy `M` là `Na`