30.

\(V_{\left(O;\dfrac{3}{4}\right)}\left(M\right)=M_1\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{3}{4}.4=3\\y_1=\dfrac{3}{4}.1=\dfrac{3}{4}\end{matrix}\right.\)

\(Q_{\left(O;-90^0\right)}\left(M_1\right)=M'\Rightarrow\left\{{}\begin{matrix}x'=y_1=\dfrac{3}{4}\\y'=-x_1=-3\end{matrix}\right.\)

\(\Rightarrow M'\left(\dfrac{3}{4};-3\right)\)

31.

\(cos\dfrac{x}{2}=-\dfrac{\sqrt{3}}{2}\Rightarrow\dfrac{x}{2}=\pm\dfrac{5\pi}{6}+k2\pi\)

\(\Rightarrow x=\pm\dfrac{5\pi}{3}+k4\pi\)

Nghiệm dương nhỏ nhất \(x=\dfrac{5\pi}{3}\Rightarrow a+b=5+3=8\)

a) (-x+31)-39=-69

   -x+31=-69+39

  -x+31=-30

-x=-30-31

-x=-61

=> x=61

b) -120-(-30-x)=-50

     -30-x=-120-(-50)

     -30-x=-70

         x=-30-(-70)

        x=40

c) 17+x- [ 352-(-400)]=-32

                 17+x-752=-32

                     17+x=-32+752

                      17+x=720

                           x=720-17

                            x=703

đúng nha

a) 2KMnO₄ +16HCl --> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Chất oxh: KMnO₄; chất khử: HCl

b) 8Al + 30HNO₃ --> 8Al(NO₃)₃ + 3N₂O + 15H₂O

31: 

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

_____0,1----------------->0,1

10FeSO₄ + 2KMnO₄ + 8H₂SO₄ --> K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

=> nKMnO₄ = 0,02 (mol)

=> \(V=\dfrac{0,02}{0,5}=0,04\left(l\right)=40\left(ml\right)\)

31) 

BTNT C: \(n_C=n_{CO_2}=n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\)

Ta có: \(m_{gi\text{ảm}}=m_{BaCO_3}-m_{CO_2}-m_{H_2O}\)

\(\Rightarrow59,1-0,3.44-m_{H_2O}=38,7\Leftrightarrow m_{H_2O}=7,2\left(g\right)\\ \Rightarrow n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,8\left(mol\right)\)

Ta có: \(n_{H_2O}>n_{CO_2}\left(0,4>0,3\right)\Rightarrow X\) thuộc dãy đồng đẳng ankan

Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

Ta có: \(n_X=n_{H_2O}-n_{CO_2}=0,1\left(mol\right)\)

Mà \(m_X=0,3.12+0,8=4,4\left(g\right)\)

\(\Rightarrow M_X=\dfrac{4,4}{0,1}=44\left(g/mol\right)\\ \Rightarrow14n+2=44\Leftrightarrow n=3\left(t/m\right)\)

Vậy X là C₃H₈ => Chọn C

c a c a b

C A C A B

\(\left\{{}\begin{matrix}\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=0\\\overrightarrow{G_0B}+\overrightarrow{G_0C}+\overrightarrow{G_0D}=0\end{matrix}\right.\)

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}\)

\(\Leftrightarrow\left(\overrightarrow{GG_0}+\overrightarrow{G_0A}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0B}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0C}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0D}\right)=\overrightarrow{0}\)

\(\Leftrightarrow4\overrightarrow{GG_0}+\overrightarrow{G_0A}+\left(\overrightarrow{G_0B}+\overrightarrow{G_0C}+\overrightarrow{G_0D}\right)=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{AG_0}=4\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{AG}+\overrightarrow{GG_0}=4\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{AG}=3\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{GA}=-3\overrightarrow{GG_0}\)

Câu 31: 

PT: \(CH_2=CH_2+H_2O\underrightarrow{t^o,xt}CH_3CH_2OH\)

→ Đáp án: A

Câu 32:

Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_X=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Gọi CTPT của X là C_nH_2n.

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_X}=5\)

Vậy: X là C₅H_10.

→ Đáp án: D

Gọi A= 1+ 2+ 3+...+ 30+ 31.

Dãy A có số các số hạng là:

( 31- 1): 1+ 1= 31( số)

Tổng dãy A là:

( 31+ 1)x 31: 2= 496.

Vậy 1+ 2+ 3+...+ 30+ 31= 496.

\(\left(31-1+1\right).\left(31+1\right):2\)

\(=31.32:2\)

\(496\)

1 + 2 + 3 +...+ 30 + 31

= [ ( 31 - 1 ) : 1 + 1] x [ ( 31 + 1 ) : 2 ]

= [ 30 : 1 + 1 ] x [ 32 : 2 ]

= 31 x 16

= 496

Học tốt !!!

Câu 9:

\(94n_{C_6H_5OH}+46n_{C_2H_5OH}=13,04\left(g\right)\left(1\right)\)

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

\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_6H_5OH}+\dfrac{1}{2}n_{C_2H_5OH}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_6H_5OH}=0,08\left(mol\right)\\n_{C_2H_5OH}=0,12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%n_{C_2H_5OH}=\dfrac{0,12}{0,12+0,08}.100\%=60\%\)

Đáp án: B

Câu 31:

Ta có: m _{tinh bột} = 1000.80% = 800 (kg) \(\Rightarrow n_{\left(C_6H_{10}O_5\right)_n}=\dfrac{800}{162n}=\dfrac{400}{81n}\left(kmol\right)\)

PT: \(\left(C_6H_{10}O_5\right)_n+nH_2O\underrightarrow{H^+,t^o}nC_6H_{12}O_6\)

\(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)

\(\Rightarrow n_{C_2H_5OH\left(LT\right)}=2n_{C_6H_{12}O_6}=2n.n_{\left(C_6H_{10}O_5\right)_n}=\dfrac{800}{81}\left(kmol\right)\)

Mà: H = 70% \(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{800}{81}.70\%=\dfrac{560}{81}\left(kmol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=\dfrac{560}{81}.46=\dfrac{25760}{81}\left(kg\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{\dfrac{25760}{81}}{0,8}=\dfrac{32200}{81}\left(l\right)\)

\(\Rightarrow V_{C_2H_5OH\left(60^o\right)}=\dfrac{\dfrac{32200}{81}}{0,6}\approx662,55\left(l\right)\)

Đáp án: A