Mỗi câu hỏi có  \(\dfrac{1}{4}\) khả năng trả lời đúng và \(\dfrac{3}{4}\) khả năng trả lời sai

Có 3 trường hợp thỏa mãn: học sinh trả lời đúng 7 câu, 8 câu, 9 câu

Xác suất:

\(P=C_9^7.\left(\dfrac{1}{4}\right)^7.\left(\dfrac{3}{4}\right)^2+C_9^8.\left(\dfrac{1}{4}\right)^8.\left(\dfrac{3}{4}\right)^1+C_9^9.\left(\dfrac{1}{4}\right)^9=\dfrac{11}{8192}\)

39B

a)

nSO2=3,36 / 22,4=0,15 mol

Cu +2H2SO4 đ -t°-> CuSO4+SO2+2H2O

=> nCu = nSO2 = CuSO4 = 0,15 mol

mCuSO4=0,15.160=24g

mZnSO4=56,2-24=32,2g

nZnSO4=nZnO= 32,2/161=0,2 mol

m=mCu+mZnO=0,15.64+0,2.81=25,8g

b)

nH2SO4 pư=2nCu+nZnO=2.0,15+0,2=0,5 mol

nH2SO4 dư=0,5.10%=0,05mol

H2SO4+BaCl2 -> BaSO4+2HCl

nH2SO4dư=nBaSO4=0,05mol

mBaSO4=0,05.233=11,65g

3b:

ĐKXĐ: \(x\in R\)

\(\sqrt{x^2+2x+4}+\left(x-1\right)\left(x+3\right)+1=0\)

=>\(\sqrt{x^2+2x+4}+x^2+2x-3+1=0\)

=>\(\sqrt{x^2+2x+4}+x^2+2x-2=0\)

=>\(x^2+2x+4+\sqrt{x^2+2x+4}-6=0\)

=>\(\left(\sqrt{x^2+2x+4}\right)^2+3\sqrt{x^2+2x+4}-2\sqrt{x^2+2x+4}-6=0\)

=>\(\left(\sqrt{x^2+2x+4}+3\right)\left(\sqrt{x^2+2x+4}-2\right)=0\)

=>\(\sqrt{x^2+2x+4}-2=0\)

=>\(\sqrt{x^2+2x+4}=2\)

=>\(x^2+2x+4=4\)

=>\(x^2+2x=0\)

=>x(x+2)=0

=>\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

a) 

P1:

\(n_{Br_2}=\dfrac{80.20\%}{160}=0,1\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

             0,1<--0,1

=> \(n_{C_2H_4\left(P_1\right)}=0,1\left(mol\right)\)

=> \(m_{C_3H_8\left(P_1\right)}=\dfrac{12,2}{2}-0,1.28=3,3\left(g\right)\)

=> \(n_{C_3H_8\left(P_1\right)}=\dfrac{3,3}{44}=0,075\left(mol\right)\)

=> \(V=\left(0,1.2+0,075,2\right).22,4=7,84\left(l\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,1+0,075}.100\%=57,143\%\\\%V_{C_3H_8}=\dfrac{0,075}{0,1+0,075}.100\%=42,857\%\end{matrix}\right.\)

b) P₂ \(\left\{{}\begin{matrix}C_2H_4:0,1\left(mol\right)\\C_3H_8:0,075\left(mol\right)\end{matrix}\right.\)

Bảo toàn C: \(n_{CO_2}=0,425\left(mol\right)\) => \(n_{BaCO_3}=0,425\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=0,5\left(mol\right)\)

Xét \(\Delta m=m_{CO_2}+m_{H_2O}-m_{BaCO_3}=0,425.44+0,5.18-0,425.197=-56,025\left(g\right)\)

=> khối lượng dd sau pư giảm 56,025 gam

B

C

B

Câu 18 : A

Câu 19 : A

Câu 20 : B

A

A

D