Theo định luật bảo toàn khối lượng có:

\(m_{Cu}+m_{O_2}=m_{CuO}\)

\(\Rightarrow m_{O_2}=m_{CuO}-m_{Cu}=10-8=2\left(g\right)\)

Chọn D

\(n_{\left(NH_4\right)_2SO_4}=1.0,2=0,2\left(mol\right)\)

PTHH:

\(2KOH+\left(NH_4\right)_2SO_4\rightarrow K_2SO_4+2H_2O+2NH_3\) 

                   0,2----------------------------------->0,4

\(V=V_{khí}=V_{NH_3}=0,4.22,4=8,96\left(l\right)\)

Nháp:

\(9\pm\sqrt{80}=9\pm4\sqrt{5}=\dfrac{72\pm32\sqrt{5}}{8}=\left(\dfrac{3\pm\sqrt{5}}{2}\right)^3\)

\(\Rightarrow \sqrt[3]{9+\sqrt{80}}=\dfrac{3+\sqrt{5}}{2}\); \(\Rightarrow \sqrt[3]{9-\sqrt{80}}=\dfrac{3-\sqrt{5}}{2}\)

\(S=\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=2+\sqrt{3}+\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\\ S=2+\sqrt{3}+3\\ S=5+\sqrt{3}\)

\(sin3x=-\dfrac{4}{3}\)

\(\Rightarrow\)\(-1\le sin3x\le1\) 

Mà \(-\dfrac{4}{3}\le-1\)

\(\Rightarrow\)PT trên vô nghiệm

\(\sqrt{\left(\sqrt{3}-1\right)^2}+\dfrac{6}{\sqrt{3}}-15\sqrt{\dfrac{1}{3}}+1\\ =\left|\sqrt{3}-1\right|+\dfrac{6}{\sqrt{3}}-15\dfrac{\sqrt{1}}{\sqrt{3}}+1\\ =\sqrt{3}-1+\dfrac{6}{\sqrt{3}}-\dfrac{15}{\sqrt{3}}+1\\ =\sqrt{3}-\dfrac{9}{\sqrt{3}}\\ =\dfrac{3}{\sqrt{3}}-\dfrac{9}{\sqrt{3}}\\ =-\dfrac{6}{\sqrt{3}}\\ =-\dfrac{\sqrt{36}}{\sqrt{3}}\\ =-\dfrac{\sqrt{3}.\sqrt{3}.\sqrt{4}}{\sqrt{3}}\\ =-\sqrt{3}.\sqrt{4}\\ =-2\sqrt{3}\)

Ta có:

\(\widehat{ABz}+\widehat{x'Bz}=180^o\) (kề bù)

\(\Leftrightarrow\widehat{ABz}+75^o=180^o\\ \Leftrightarrow\widehat{ABz}=180^o-75^o=105^o\)

Mà \(\widehat{yAx'}=105^o=\widehat{ABz}\)

\(\Rightarrow Ay//Bz\) (2 góc trên bằng nhau ở vị trí so le trong)

hay \(yy'//Bz\) (A nằm trên đoạn \(yy'\))

Cách 2:

Ta có:

\(\widehat{BAy'}+\widehat{yAB}=180^o\) (kề bù)

\(\Leftrightarrow\)\(\widehat{BAy'}=180^o-\widehat{yAB}\) 

\(\Leftrightarrow\widehat{BAy'}=180^o-105^o=75^o\)

Mà \(\widehat{x'Bz}=75^o=\widehat{BAy'}\)

\(\Rightarrow Bz//Ay'\) (2 góc trên bằng nhau ở vị trí đồng vị)

hay \(Bz//yy'\) (A nằm trên đoạn \(yy'\))

\(n_{H_2}=\dfrac{1,4874}{24,9}=0,06\left(mol\right)\)

BTNT H:

 \(2n_{H_2}=n_{HCl}\\ \Rightarrow n_{HCl}=2.0,06=0,12\left(mol\right)\)

BTKL có: \(m_{kl}+m_{HCl}=m_{muối}+m_{H_2}\)

=> \(m_{muối}=1,58+0,12.36,5-0,06.2=5,84\left(g\right)\)

làm lại đây, xin lỗi nhé: )