Giải giúp em bài này với ạ em cảm ơn 
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TT
2
TT
2
DT
Đỗ Thanh Hải
CTVVIP
24 tháng 5 2021
1 The meeting was canceled 3 days ago
2 She told me she was watching a film with her sister then
3 I admire the guitarist who is perfroming on the stage
4 Had it not been for Pauline's interest, the project would have been abandoned
KY
3 tháng 8 2021
He asked me to know what the time was.
She asked me when we would meet again.
She asked him If he was crazy.
He asked me to know when they had lived.
He asked her If she would be at the party.
She asked me If I could meet her at the station.
The teacher asked me to know who knew the answer
She asked him to know why he didn't help her.
He asked me If I had seen that car.
The mother asked the twins If they had tidied up their room.
Tom asked how much that computer was
LT
3 tháng 8 2021
11. what the time was.
12. when they would meet again.
13. if he was crazy.
14. where they had lived.
15. if she would be at the party.
16. if I could meet her at the station.
17. who knew the answer.
18. why I didn't help her.
19. if I had seen that car.
20. if they had tidied up their room.
Bài 3:
asked how much that computer was.
19 tháng 9 2021
Ta có: \(\left\{{}\begin{matrix}p+e+n=52\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=17\\n=18\end{matrix}\right.\)
KN
1
S
2 tháng 4 2022
Câu 2 :
a) PTHH : \(2Al+3S-->Al_2S_3\)
\(Zn+S-->ZnS\)
mS = 29,55 - 15,15 = 14,4 (g)
b) nS = 14,4/32 = 0,45 (mol)
PTHH : \(Al_2S_3+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2S\)
\(ZnS+H_2SO_4-->ZnSO_4+H_2S\)
\(H_2S+2AgNO_3-->Ag_2S+2HNO_3\)
Bảo toàn S : nAg2S = nH2S = nS = 0,45 (mol)
=> mktAg2S = 111,6 (g)
c) PTHH : \(H_2S+CuCl_2-->CuS+2HCl\)
nH2S = 0,45/2 = 0,225 (mol)
Theo pthh : nCuS = nH2S = 0,225 (mol)
=> mCuS = 0,225.96 = 21,6 (g)
25 tháng 3 2022
a)
Gọi số mol Fe, Al là a, b (mol)
=> 56a + 27b = 8,94 (1)
\(n_{H_2}=\dfrac{0,36}{2}=0,18\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
a------------->a----->a
2Al + 6HCl --> 2AlCl3 + 3H2
b---------------->b------>1,5b
=> a + 1,5b = 0,18 (2)
(1)(2) => a = 0,15 (mol); b = 0,02 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{8,94}.100\%=93,96\%\\\%m_{Al}=\dfrac{0,02.27}{8,94}.100\%=6,04\%\end{matrix}\right.\)
b)
mFeCl2 = 0,15.127 = 19,05 (g)
mAlCl3 = 0,02.133,5 = 2,67 (g)
25 tháng 3 2022
Gọi số mol Mg, Zn là a, b (mol)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a---------------->a----->a
Zn + 2HCl --> ZnCl2 + H2
b--------------->b---->b
=> a + b = 0,3 (1)
Và 95a + 136b = 36,7 (2)
(1)(2) => a = 0,1 (mol); b = 0,2 (mol)
\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{0,1.24+0,2.65}.100\%=15,6\%\\\%m_{Zn}=\dfrac{0,2.65}{0,1.24+0,2.65}.100\%=84,4\%\end{matrix}\right.\)
30 tháng 3 2022
a, Hiện tượng: mất dần màu vàng lục của Clo, có khí không màu thoát ra
\(K_2CO_3+Cl_2\rightarrow KCl+KClO+CO_2\uparrow\)
b, \(F_2+H_2\xrightarrow[\text{nhiệt độ âm}]{\text{bóng tối}}2HF\)
\(Cl_2+H_2\underrightarrow{as}2HCl\)
\(Br_2+H_2\underrightarrow{t^o}2HBr\\ I_2+H_2\underrightarrow{t^o,xt}2HI\)
a, \(n_{HCl}=0,1.1,5=0,15\left(mol\right);n_{H_2}=\dfrac{1,288}{22,4}=0,0575\left(mol\right)\)
Ta có: \(\dfrac{0,15}{8}>\dfrac{0,0575}{4}\) ⇒ HCl dư
⇒ tính số mol theo H2
Fe + 2HCl ----> FeCl2 + H2
x x x x
2Al + 6HCl ----> 2AlCl3 + 3H2
y 3y 2y 1,5y
Ta có: hệ pt: \(\left\{{}\begin{matrix}56x+27y=1,795\\x+1,5y=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,025\left(mol\right)\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,02.56.100\%}{1,795}=62,4\%;\%m_{Al}=100\%-62,4\%=37,6\%\)
b, Trong dd X có HCl dư, FeCl2 và AlCl3
\(n_{HCldư}=0,15-0,02-3.0,025=0,055\left(mol\right)\)
HCl + KOH ----> KCl + H2O
0,055 0,055
2KOH + FeCl2 ----> Fe(OH)2 + 2KCl
0,04 0,02
3KOH + AlCl3 -----> Al(OH)3 + 3KCl
0,15 0,05
\(n_{KOH}=0,055+0,04+0,15=0,245\left(mol\right)\)
\(\Rightarrow V_{KOH}=\dfrac{0,245}{1}=0,245\left(l\right)=245\left(ml\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a a a
2Al + 6HCl ---> 2AlCl3 + 3H2
b b 1,5b
\(n_{H_2}=\dfrac{1,228}{22,4}=0,0575\left(mol\right)\)
Hệ pt \(\left\{{}\begin{matrix}56a+27b=1,795\\a+1,5b=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,02\left(mol\right)\\b=0,025\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,02.56=1,12\left(g\right)\\\%m_{Al}=0,025.27=0,675\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{1,12}{1,795}=62,4\%\\\%m_{Al}=100\%-62,4\%=37,6\%\end{matrix}\right.\)
PTHH:
FeCl2 + 2KOH ---> Fe(OH)2 + 2KCl
0,02 0,04 0,02
AlCl3 + 3KOH ---> Al(OH)3 + 3KCl
0,025 0,075 0,025
\(\rightarrow V_{ddKOH}=\dfrac{0,075+0,04}{1}=0,115\left(l\right)\)