d b c b d

25.

\(n_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi CTTQ của hai ankan đồng đăng kế tiếp là: \(C_{\overline{n}}H_{2\overline{n}+2}\)

=> \(\overline{M}_{hh}=\dfrac{24,8}{0,5}=49,6\)

Có: \(14\overline{n}+2=49,6\)

=> \(\overline{n}=3,4\) (với đk \(\overline{n}\) nguyên dương)

=> 2 ankan lần lượt là \(C_3H_8,C_4H_{10}\)

Gọi x: n \(C_3H_8\)

       y: \(n_{C_4H_{10}}\)

có:  

x + y = 0,5

44x + 58y = 24,8

=> x = 0,3

y = 0,2

=> \(\%_{V_{C_3H_8}}=\dfrac{0,3.22,4.100}{11,2}=60\%\)

=> \(\%_{V_{C_4H_{10}}}=100-60=40\%\)

Vậy chọn C.

$n_{Al(NO_3)_3} = n_{Al} = 0,28(mol)$
$n_{Zn(NO_3)_2} = n_{Zn} = 0,18(mol)$

$\Rightarrow m_{NH_4NO_3} = m_{muối} - m_{Al(NO_3)_3} - m_{Zn(NO_3)_2} = 99,66 - 0,28.213 - 0,18.189 = 6(gam)$

$\Rightarrow n_{NH_4NO_3} = 0,075(mol)$

Bảo toàn electron : 

$3n_{Al} + 2n_{Zn} = 8n_{NH_4NO_3} + 3n_{NO}$

$\Rightarrow n_{NO} = 0,2(mol)$

Đáp án B

22) \(n_{C_3H_6}=\dfrac{2,52}{42}=0,06\left(mol\right)\)

PTHH: \(2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\)

             0,06------------->0,18---->0,18

\(\Rightarrow m_{t\text{ăng}}=0,18.\left(44+18\right)=11,16\left(g\right)\)

Chọn B

23) Propan là \(CH_3-CH_2-CH_3\). Chọn D

24) Chất không tạo kết tủa là \(CH_3-C\equiv C-CH_3\) do không phải là ank-1-in

\(CH\equiv CH+2AgNO_3+2NH_3\rightarrow CAg\equiv CAg\downarrow+2NH_4NO_3\\ CH_3-C\equiv CH+AgNO_3+NH_3\rightarrow CH_3-C\equiv CAg\downarrow+NH_4NO_3\\ CH\equiv C-CH=CH_2+AgNO_3+NH_3\rightarrow CAg\equiv C-CH=CH_2\downarrow+NH_4NO_3\)

Chọn A

25) BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,9\left(mol\right)\\n_H=2n_{H_2O}=1,2\left(mol\right)\end{matrix}\right.\Rightarrow m=0,9.12+1,2=12\left(g\right)\)

Chọn C

26) \(CH_4+4Cl_2\xrightarrow[]{a/s}CCl_4+4HCl\)

Chọn D

Có đáp số 6(mJ) ko bạn?

#\(Errink \times Cream\)

#\(yGLinh\)

Câu 27:

\(C_3H_4+AgNO_3+NH_3\rightarrow C_3H_3Ag+NH_4NO_3\\ C_2H_2+2AgNO_3+2NH_3\rightarrow Ag_2C_2+2NH_4NO_3\\ n_{C_3H_4}=\dfrac{1,8}{40}=0,045\left(mol\right);n_{C_2H_2}=\dfrac{0,52}{26}=0,02\left(mol\right)\\ n_{C_3H_3Ag}=n_{C_3H_4}=0,045\left(mol\right)\\ n_{Ag_2C_2}=n_{C_2H_2}=0,02\left(mol\right)\\ m_{kt}=m_{C_3H_3Ag}+m_{Ag_2C_2}=0,045.147+240.0,02=11,415\left(g\right)\\ Chọn.B\)

\(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\\ Đặt:n_{C_3H_6}=a\left(mol\right);n_{C_3H_4}=b\left(mol\right)\\ PTHH:C_3H_6+Br_2\rightarrow C_3H_6Br_2\\ C_3H_4+2Br_2\rightarrow C_3H_4Br_4\\ \Rightarrow\left\{{}\begin{matrix}42a+40b=6,2\\a+2b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{C_3H_6}=\dfrac{0,1.42}{6,2}.100\approx67,74\%\\ Chọn.A\)

\(n_{Br_2}=\dfrac{0,8}{160}=0,005\left(mol\right)\\ C_2H_4\left(dư\right)+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4\left(dư\right)}=n_{Br_2}=0,005\left(mol\right)\\ n_{C_2H_4\left(đã.p.ứ.với.H_2\right)}=\dfrac{2,24}{22,4}-0,005=0,095\left(mol\right)\\ C_2H_4+H_2\rightarrow\left(Ni,t^o\right)C_2H_6\\ Vì:\dfrac{0,095}{1}< \dfrac{3,36:22,4}{1}\Rightarrow C_2H_4hết,H_2dư\\ H=\dfrac{0,095}{0,1}.100\%=95\%\\ Chọn.C\)

10. a present was sent to me last week
11. More information was given to us by her 

Câu 23: C

Câu 24: C

a