giúp mik bài 5,6 vs ạ
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5. a) Fe + H2SO4 --------> FeSO4 + H2
Mg + H2SO4 --------> MgSO4 + H2
Gọi x, y lần lượt là số mol của Fe, Mg
Ta có \(\left\{{}\begin{matrix}56x+24y=12\\x+y=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,15\\y=0,15\end{matrix}\right.\)
=> \(\%m_{Fe}=\dfrac{0,15.56}{12}.100=70\%\)
=> %mMg = 100 -70 = 30%
b) Ta có n H2SO4 = n Fe + nMg = 0,15 + 0,15 =0,3 (mol)
=>a = \(CM_{H_2SO_4}=\dfrac{0,3}{0,2}=1,5M\)
c) \(CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\)
\(CM_{MgSO_4}=\dfrac{0,15}{0,2}=0,75M\)
6. a) 2Al + 6HCl → 2AlCl3 + 3H2
Zn + 2HCl →ZnCl2 + H2
Gọi x, y lần lượt là số mol của Al, Zn
\(\left\{{}\begin{matrix}27x+65y=11,9\\\dfrac{3}{2}x+y=\dfrac{8,96}{22,4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
=> \(\%m_{Al}=\dfrac{27.0,2}{11,9}.100=45,38\%\)
=> %m Zn = 100- 45,38=54,62%
b) \(a=C\%_{HCl}=\dfrac{\left(0,2.2+0,1.2\right).36,5}{200}.100=10,95\%\)
c) \(m_{ddsaupu}=11,9+200-\left(0,2.2+0,1.2\right)=211,3\left(g\right)\)
=> \(C\%_{AlCl_3}=\dfrac{0,2.133,5}{211,3}.100=12,64\%\)
\(C\%_{ZnCl_2}=\dfrac{0,1.136}{211,3}.100=6,44\%\)
câu 6 thể tích 1kg khí ở 0 độ 1,29 m3
áp dụng đẳng quá trình \(\dfrac{P_0V_0}{T_0}=\dfrac{PV}{T}\Leftrightarrow\dfrac{1,01.10^5.1,29}{0+273}=\dfrac{3.10^5.V}{70+273}\Rightarrow V\approx0,5456\left(m^3\right)\)
khối lượng riêng 0,5456kg/m3
$n_{CaO} = \dfrac{5,6}{56} =0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
1 interested in travelling by plane
2 15 minutes riding his bike to school
3 is made by my mother everymorning
4 take care of her little brother
5 able to speak 2 languages when he was young
6 able to play the piano
7 is played all over VN
8 to play voleyball very well when he was young
9 to go fishing when he was young
10 way to the shopping mall
Bài 5: