Câu 33:

\(n_{C_6H_2Br_3\left(OH\right)}=\dfrac{4,965}{331}=0,015\left(mol\right)\)

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3\left(OH\right)+3HBr\)

Theo PT: \(n_{Br_2}=3n_{C_6H_2Br_3\left(OH\right)}=0,045\left(mol\right)\)

\(\Rightarrow m_{Br_2}=0,045.160=7,2\left(g\right)\)

Đáp án: D

Câu 34:

Ta có: 94n_C6H5OH + 32n_CH3OH = 15,8 (1)

PT: \(C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\)

\(CH_3OH+Na\rightarrow CH_3ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_6H_5ONa}+\dfrac{1}{2}n_{CH_3OH}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_6H_5OH}=0,1\left(mol\right)\\n_{CH_3OH}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%n_{CH_3OH}=\dfrac{0,2}{0,1+0,2}.100\%\approx66,67\%\)

Đáp án: C

10: B

33: B

34: D

35: D

33B

\(\frac{5}{22}\cdot\frac{33}{17}+\frac{15\cdot101}{34}-\frac{30}{34}=\frac{5\cdot11\cdot3}{17\cdot11\cdot2}+\frac{15\cdot101-15\cdot2}{34}=\frac{15}{34}+\frac{15.99}{34}\)

=100*15/34=1500/34 (bấm máy tính) = 44.1176471....

\(=\dfrac{33}{17}\cdot\dfrac{34}{3}\cdot\dfrac{2}{5}=\dfrac{2\cdot11\cdot2}{5}=\dfrac{44}{5}\)

\(\dfrac{33}{17}\times\dfrac{34}{3}\times\dfrac{2}{5}=11\times2\times\dfrac{2}{5}=\dfrac{44}{5}\)

44/5

Ta có 

F=[(123-17)-(123+33)]- {34-[34+(57-50)-7]}

F=(123-17-123-33)-(34-34+7-7)

F=(0-50)-0+=O

F=-50+