Câu 27:

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

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

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

3

26 C

27 B

3

PHAN TRAN HUYEN TRANG ơi bạn lầm đề rồi!

26/81+ 4/27=38/81

26/81+4/27=38/81

mình trả lời trước nhé

nếu đúng mick nhé

oe ít bài?

bị hack

ko thấy bài, lỗi

\(\frac{x}{27}-\frac{2}{9}=\frac{6}{18}\)

\(\frac{x}{27}=\frac{6}{18}-\frac{2}{9}\)

\(\frac{x}{27}=\frac{3}{9}-\frac{2}{9}\)

\(\frac{x}{27}=\frac{1}{9}\)

\(\frac{x}{27}=\frac{3}{27}\)

\(x=3\)

Ấy nhầm, làm lại nha:

\(\frac{x}{27}-\frac{2}{9}=\frac{6}{18}\)

\(\frac{x}{27}=\frac{6}{18}+\frac{2}{9}\)

\(\frac{x}{27}=\frac{3}{9}+\frac{2}{9}\)

\(\frac{x}{27}=\frac{5}{9}\)

\(\frac{x}{27}=\frac{15}{27}\)

\(x=15\)

\(\frac{15}{26}+\frac{x}{13}=\frac{46}{52}\)

\(\frac{x}{13}=\frac{46}{52}-\frac{15}{26}\)

\(\frac{x}{13}=\frac{23}{26}-\frac{15}{26}\)

\(\frac{x}{13}=\frac{8}{26}\)

\(\frac{x}{13}=\frac{4}{13}\)

\(x=4\)

Xét vế phải :

\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)