Giúp em câu 2,3 với ạ
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Wlk=(Z*mp-(A-Z)*mn-mCo)*c^2
=(27*1,0073u +31*1,0087u-56,9540u)*c^2=1,5128u*c^2=1,5128*931,5
=1409,1732MeV
--->Wr=Wlk/A=1409,1732/58=24,296MeV
Bài 1:
c: Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là
x1=1; \(x2=\dfrac{c}{a}=\dfrac{3\sqrt{2}+1}{1-\sqrt{2}}\)
a: \(\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\3x+12y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-6y=-2\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Câu 2:
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2a 3a a 3a
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b b b b
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25mol\)
Ta có: \(\left\{{}\begin{matrix}54a+65b=9.2\\3a+b=0.25\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.05\\b=0.1\end{matrix}\right.\)
a.\(\%m_{Al}=\dfrac{0.05\times54\times100}{9.2}=29.3\%\)
\(\%m_{Zn}=100-29.3=70.7\%\)
Vdd sau phản ứng = 9.2 + 600 - 0.0056 = 609.2ml
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0.05}{0.6092}=0.08M\)
\(CM_{ZnSO_4}=\dfrac{0.1}{0.6092}=0.16M\)
Câu 3:
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0.2 0.2 0.2 0.2
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
\(n_{H_2}=\dfrac{4.48}{22.4}=0.2mol\)
a. \(\%m_{Mg}=\dfrac{0.2\times24\times100}{12}=40g\)
\(\%m_{FeO}=100-40=60\%\)
b. \(n_{FeO}=\dfrac{12-0.2\times24}{72}=0.1mol\)
m muối khan \(=m_{MgSO_4}+m_{FeSO_4}=0.2\times120+0.1\times152=39.2g\)
\(C2:y=ax^2+bx+c\) \(\left(P\right)\) \(đi\) \(qua\)\(M\left(1;5\right)vàN\left(-2;8\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+2=5\\4a-2b+2=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow S=3a+4b+c=3.2+4+2=12\)
\(C3a,\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}\left(2;-4\right)\\\overrightarrow{AC}\left(2;-1\right)\end{matrix}\right.\)
\(b,D\left(xo;yo\right)\Rightarrow ABCD\) \(là\) \(hbh\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\Leftrightarrow\left\{{}\begin{matrix}3-xo=2\\1-yo=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xo=1\\yo=5\end{matrix}\right.\) \(\Rightarrow D\left(1;5\right)\)
Luôn cả câu 4 nữa ạ