Question

Answer 1

Câu 7 : 

\(K_2O+H_2O\rightarrow2KOH\) : Hóa hợp 

\(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\) : Thế 

\(H_2+\dfrac{1}{2}O_2\underrightarrow{^{^{t^0}}}H_2O\) : Hóa hợp 

\(KNO_3\underrightarrow{^{^{t^0}}}KNO_2+\dfrac{1}{2}O_2\) : Phân hủy

 

Answer 2

Câu 8 : 

\(a.\) \(HCl\) : Axit 

\(b.Al_2O_3\) : Oxit lưỡng tính

\(c.Fe\left(OH\right)_3\) : Bazo 

\(d.Ca_3\left(PO_4\right)_2\) : Muối

Answer 3

Câu 9 : 

\(n_{CuO}=\dfrac{4}{80}=0.05\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(0.05....0.05.....0.05\)

\(m_{Cu}=0.05\cdot64=3.2\left(g\right)\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

Answer 4

Câu 7:

a, \(K_2O+H_2O\rightarrow2KOH\)

b, \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)

c, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

d, \(2KNO_3\underrightarrow{t^o}2KNO_2+O_2\)

Câu 8:

a, HCl - axit

b, Al₂O₃ - oxit 

c, Fe(OH)₃ - bazơ

d, Ca₃(PO₄)₂ - muối

Câu 9:

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

______0,05_0,05_0,05 (mol)

b, m_Cu = 0,05.64 = 3,2 (g)

c, V_H2 = 0,05.22,4 = 1,12 (l)

Bạn tham khảo nhé!