Question

Giúp mình vs ạ

Answer 1

Câu 1 : 

\(n_{CO_2}=\dfrac{2.10^{23}}{6.10^{23}}=0,33\left(mol\right)\)

Answer 2

Cau 2:

1 . 4Na + O₂ -> 2Na₂O

2. BaCl₂ + H₂SO₄ -> BaSO₄ + 2HCl

3. 2Al(OH)₃ -> Al₂O₃ + 3H₂O

 

Answer 3

5)

\(a.4K+O_2\underrightarrow{t^o}2K_2O\\ b.BaCl_2+2AgNO_3\rightarrow2AgCl+Ba\left(NO_3\right)_2\\ c.6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)

Answer 4

1) \(n_{CO_2}=\dfrac{2.10^{23}}{6.10^{23}}=\dfrac{1}{3}\left(mol\right)\)

2)

\(4Na+O_2\underrightarrow{t^o}2Na_2O\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

 

 

Answer 5

Câu 4 : 

\(n_{H_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

Answer 6

Câu 3 :

a) 2Zn + O₂ -> 2ZnO

b) \(m_{Zn}+m_{O_2}=m_{ZnO}\)

c) \(m_{O_2}=28-15=13\left(g\right)\)