C35: B

C36: C

C37: D

C38: D

C39: A

C40: B

C41: A

C42: A

C43: A

C44: B

C45: C

C46: D

C47: B

C48: C

C49: D

C50: D

C12: B

C13: C

C14: C

C15: A

C16: B

C17: C

C18: C

C19: D

C20: B

C21: D

C22: D

C23: C

C24: A

C25: B

C26: D

C27: C

C28: D

C29: A

C30: D

C31: D

C32: B

C33: A

C2: A

C3: C

C4: C

C5: A

C6: A

C7: A

C8: C

C9: D

C10: B

\(\Rightarrow X_2=66-24=42\)

\(\Rightarrow X=21\)

\(\Rightarrow2p_X+n_X=21\left(1\right)\)

\(4p_X+16-2n_X-8=22\)

\(\Leftrightarrow4p_X-2n_X=14\)

\(\Rightarrow2p_X-n_X=7\left(2\right)\)

Từ 1, 2 suy ra :

\(4p_X=28\)

\(\Rightarrow p_X=7\)

-> Nito (N)

-> N₂O 

Gọi x là % B10; %B11 là (100-x) 

\(\overline{A}=10,81=\dfrac{10+11\left(100-x\right)}{100}\)

\(\Rightarrow x=19\)

\(m_{B10}=19\%.2,162=0,41078\left(g\right)\)

\(n_{B10\left(Boron\right)}=\dfrac{0,41078}{11}\approx0,03734\left(mol\right)\)

\(N=0,03734.6,022.10^{23}=2,25.10^{22}\left(ng.tử\right)\)

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Gọi M là hh 2 kim loại ở 2 chu kì liên tiếp

\(M+2HCl\rightarrow MCl_2+H_2\)

\(n_M=n_{H2}=0,15\left(mol\right)\)

\(M_M=\dfrac{4,4}{0,15}=29,33\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow M_{Mg}=24< 29,33< M_{Ca}=40\)

Vậy 2 kim loại cần tìm là Mg va Ca