\(BaCl_2+H_2SO_4\xrightarrow[]{}BaSO_4\left(kết.tủa.trắng\right)+HCl\\ Ba\left(NO_3\right)_2+H_2SO_4\xrightarrow[]{}BaSO_4\left(kết.tủa.trắng\right)+2HNO_3\)

chọn A

A nha

\(Na_2O+H_2O->2NaOH\\ n_{Na_2O}=\dfrac{5,6}{62}=0,0903\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,0903\cdot2}{2}=0,0903\left(M\right)\)

Na₂O + H₂O --> 2NaOH

n_Na₂O = 5,6/62 = 0,09 (mol)

=> n_NaOH = 0,18 (mol)

=> C_M NaOH = 0,18/2 = 0,08 (M)

\(TN1:\\ n_{H_2SO_4}=\dfrac{2x+y}{3}\cdot0,01\left(mol\right)\\ Zn+H_2SO_4->ZnSO_4+H_2\\ \dfrac{2x+y}{3}\cdot0,01=\dfrac{0,05824}{22,4}\left(I\right)\\ TN2:\\ n_{H_2SO_4}=\dfrac{x+3y}{4}\cdot0,05\\ H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ 0,1\cdot\dfrac{x+3y}{4}=0,05\cdot16,8:40\left(II\right)\\ \Rightarrow x=0,3\left(M\right);y=0,18\left(M\right)\)

\(TN1:\\ n_{H_2SO_4}=\dfrac{2x+y}{3}\cdot0,01\left(mol\right)\\ Zn+H_2SO_4->ZnSO_4+H_2\\ \dfrac{2x+y}{3}\cdot0,01=\dfrac{0,05824}{22,4}\left(I\right)\\ TN2:\\ n_{H_2SO_4}=\dfrac{x+3y}{4}\cdot0,05\\ H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ 0,1\cdot\dfrac{x+3y}{4}=0,05\cdot16,8:40\left(II\right)\\ \Rightarrow x=0,3\left(M\right);y=0,18\left(M\right)\)

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1 \left(mol\right)\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,1 -----------------> 0,1

\(CM_{base}=CM_{NaOH}=\dfrac{0,1}{0,2}=0,5M\)

b

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

0,05 <------ 0,1

\(V_{H_2SO_4}=\dfrac{0,05}{0,2}=0,25\left(l\right)\Rightarrow V_{dd.H_2SO_4}=\dfrac{0,25.100}{20}=1,25\left(l\right)\)

a, Sao 80ml lại là 10% bạn nhỉ ?

\(b,n=C_M.V=0,5.0,8=0,4\left(mol\right)\)

\(m_{NaOH}=0,4.40=16\left(g\right)\)

Đề thiếu dữ kiện 

9,61 lít hay gam

Tham Khảo:

\(1.=15x^3-6x^2-3x\)

\(b.\left(2y^4-1,5y^3-3,5y^2+4y\right)\)

\(c.-4x^4+2x^3-x^2\)

\(d.x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

Bạn tách câu hỏi ra rôi gửi vào đúng môn nhé 

\(C_{\%}=\dfrac{87,5}{87,5+350}.100\%=20\%\)

\(a.MgCO_3-t^{^0}->MgO+CO_2\\ CaCO_3-t^{^0}->CaO+CO_2\\ b.n_{MgCO_3}=a,n_{CaCO_3}=b\\ 84a+100b=26,8\\ 40a+56b=13,6\\ a=0,2;b=0,1\\ \%m_{MgCO_3}=\dfrac{84a}{26,8}.100\%=62,69\%\\ \%m_{CaCO_3}=37,31\%\\ c.V_{CO_2}=22,4\left(a+b\right)=6,72L\)