Xét \(2\Delta:\Delta APC\) và \(\Delta BQA\) có:

\(\left\{{}\begin{matrix}\widehat{APC}=\widehat{BQA}=90^o\\\widehat{BAQ}=\widehat{ACP}\left(slt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta APC\sim\Delta BQA\left(g-g\right)\)

\(x:\dfrac{3}{8}=12\)

\(x=12\times\dfrac{3}{8}=\dfrac{9}{2}\)

1.a. \(pthh:Mg+2HCl--->MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo pt: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)

2. 

\(Cu+2H_2SO_{4_đ}\overset{t^o}{--->}CuSO_4+SO_2+2H_2O\)

\(BaCl_2+H_2SO_4--->BaSO_4\downarrow+HCl\)

\(2P+5H_2SO_{4_đ}\overset{t^o}{--->}2H_3PO_4+5SO_2+2H_2O\)

\(SO_2+Br_2+2H_2O--->H_2SO_4+2HBr\)

-1 và 1 cx là ước của -7 em nhé

3/5 = 9/15 nên ko phải C đâu nhé

Câu 17: A

Câu 18: D

B

\(y\times\dfrac{1}{3}+y\times\dfrac{2}{5}=\dfrac{11}{15}\)

\(y\times\left(\dfrac{1}{3}+\dfrac{2}{5}\right)=\dfrac{11}{15}\)

\(y\times\dfrac{11}{15}=\dfrac{11}{15}\)

\(y=1\)

BaO + H₂O ---> Ba(OH)₂

Na₂O + H₂O ---> 2NaOH

a. \(BaO+H_2O--->Ba\left(OH\right)_2\)

b. \(Na_2O+H_2O--->2NaOH\)