Giúp mik bài 7,8 vs. Mik cám ơn
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Bài 3:
Ta có: a//b
nên \(x+y=180\)
mà \(2x-3y=0\)
nên \(\left\{{}\begin{matrix}x+y=180\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=180\\2x-3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=180\\x+y=180\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=36\\x=144\end{matrix}\right.\)
C
1 cooking
2 listening
3 reading
4 playing
5 cycling
1 People collect a lot of things such as stamps, milk bottle labels, comic books as well as car toys
2 She is 20 years old
3 She is a hairdresser
4 It is collecting candy shells
5 Her friends and relatives
B
1 F
2 F
3 F
4 F
5 T
`a)1001^2`
`=(1000+1)^2=1000000+2000+1`
`=1002001`
`b)29,9.30,1`
`=(30-0,1)(30+0,1)`
`=30^2-0,1^2`
`=900-0,01=899,99`
`c)199^2=(200-1)^2`
`=40000-400+1`
`=39601`
`d)84^2-16^2`
`=(84-16)(84+16)`
`=100.68`
`=6800`
`e)313^2-312^2`
`=(313-312)(313+312)`
`=625`
`f)47.53`
`=(50-3)(50+3)`
`=2500-9=2491`
Bài làm
x - ( 5 - x ) = x - 15
=> x - 5 + x - x + 15 = 0
=> 10 - x = 0
=> x = 10
Vậy x = 10
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,1____0,3_____0,1_____0,15 (mol)
\(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
0,1______0,3__________0,05____0,15_____0,3 (mol)
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
0,15_____0,3________0,15___0,15_____0,3 (mol)
Ta có: \(m_{Al}+m_{Cu}=0,1\cdot27+0,15\cdot64=12,3\left(g\right)\)
Bài 7:
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)=n_{Mg}=n_{H_2}=n_{MgSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Ag}=15,6-0,2\cdot24=10,8\left(g\right)\\V_{H_2}=0,2\cdot22,4=2,24\left(l\right)\\C_{M_{MgSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)