Question

cho mik xin đáp án với ạ

Answer 1

C1: B

C2: A

C3: C

C4: B

C5: B

C6: B

PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Ta có: \(n_R=\dfrac{9,6}{M_R}\left(mol\right)\)

\(n_{RO}=\dfrac{16}{M_R+16}\left(mol\right)\)

Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{9,6}{M_R}=\dfrac{16}{M_R+16}\Rightarrow M_R=24\left(g/mol\right)\)

→ R là Mg.

C8: Ta có: \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

____0,1____________0,1____0,05 (mol)

⇒ m_KOH = 0,1.56 = 5,6 (g)

V_H2 = 0,05.22,4 = 1,12 (l)

C9: A

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\)

C10: A