Đặt \(m_{ankan}=100g\)

\(M_Y=2.14,5=29\)

\(\rightarrow n_Y=\frac{100}{29}mol\)

\(Ankan\rightarrow Ankan'+Anken\)

\(Ankan\rightarrow Anken+H_2\)

\(\rightarrow\text{Σ}n_{SP}=2n_{thamgia}\)

\(\rightarrow n_{crakingthamgia}=\frac{100}{29}mol\)

\(\rightarrow n_{ankanthamgia}=\frac{50}{29}mol\)

\(\rightarrow M_{ankan}=\frac{100}{\frac{50}{29}}=58g/mol\)

Vậy Ankan là \(C_4H_{10}\)

 công thức phân tử C10H14N2.

1.     Kim takes hẻ dog for a ưalk in the evenings.

Put the verbs in brackets into the correct form.

1.      Kim _takes_________ (take) her dog for a walk in the evenings. 

2.      Call later. They ___studies__________ (study) for their exam now. 

3.      How much __cost the books___________ (the book/ cost)? 

4.      Take an umbrella. It __raining___________ (rain) at the moment.

5.      Why don’t we _____bought____ (buy) those pairs of shoes? 

6.      Hey! You must not __runs_________ (run) in that area! 

7.      Who ____is be___ (be) your favorite MC?

8.      Min and Anne are fond of _____drawing________ (draw) and __making___________ (make) origami.

9.      It takes her 10 minutes _is make_________ (make) this model.

10.  Would you __goes___________ (go) to the cinema with me tonight?

TA lớp 7 còn dễ hơn cả lớp 6

@ Lê Thanh Thảo sĩ ít thôi

a. PTHH: \(KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

b. \(H=100\%\)

\(n_{KMnO_4}=\frac{3,6}{158}=0,023mol\)

Theo phương trình \(n_{O_2}=0,5n_{KMnO_4}=0,046mol\)

\(\rightarrow V_{O_2}=0,0115.22,4.100\%=0,2576l\)

c. H = 80%

\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

a. \(n_{Fe}=\frac{m}{M}=\frac{5,6}{56}=0,1mol\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ban đầu: 0,1           0,1                                    mol

Trong pứng:  0,05     0,1                0,05           mol

Sau pứng:      0,05     0                  0,05           mol

\(\rightarrow n_{H_2}=n_{Fe}=0,1mol\)

\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,1.22,4=2,24l\)

b. \(n_{Fe}=\frac{m}{M}=\frac{5,6}{56}=0,1mol\)

\(\rightarrow n_{H_2}=n_{HCl}=\frac{0,1.1}{2}=0,05mol\)

\(\rightarrow V_{H_2}=n.22,4=0,05.22,4=1,12l\)

a. PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Ban đầu:   0,1             0,2                                   mol

Trong pứng:  0,1       0,1               0,1             0,1     mol

Sau pứng:      0         0,1               01,             0,1     mol

b. \(n_{Zn}=\frac{m}{M}=\frac{6,5}{65}=0,1mol\)

\(100ml=0,1l\)

\(n_{H_2SO_4}=C_M.V=2.0,1=0,2mol\)

\(\rightarrow n_{H_2}=n_{Zn}=0,1mol\)

\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,1.22,4=2,24l\)

c. \(V_{sau}=V_{H_2SO_4}=0,1l\)

\(\rightarrow C_{M_{H_2SO_4\left(dư\right)}}=\frac{n}{V_{sau}}=\frac{0,1}{0,1}=1M\)

Theo phương trình \(n_{ZnSO_4}=n_{Zn}=0,1mol\)

\(\rightarrow C_{M_{ZnSO_4}}=\frac{n}{V_{sau}}=\frac{0,1}{0,1}=1M\)