\(CT:M_2O\)

\(\%O=\dfrac{16}{2M+16}\cdot100\%=25.8\%\)

\(\Leftrightarrow M=23\)

\(M:Na\left(Natri\right)\)

cảm ơn ạ

Đặt A = 1+2+2^2+2^3+....+2^60

2A = 2+2^2+2^3+2^4+.....+2^61

2A-A= ( 2+2^2+2^3+....+2^61)-(1+2+2^2+.....+2^60)

A = 2^61-1

http://olm.vn/hoi-dap/question/232902.html

Câu 7:

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)

c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)

Câu 8:

a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)

c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

\(\frac{1}{12}-\left(-\frac{1}{6}-\frac{1}{4}\right)\)

\(=\frac{1}{12}-\left(-\frac{2}{12}-\frac{3}{12}\right)\)

\(=\frac{1}{12}+\frac{2}{12}+\frac{3}{12}\)

\(=\frac{1}{2}\)

Thanks bạn cute Jeon Koo Koo nhìu nha , tớ cảm ơn pạn rất nhìu :3

Câu 1:

a; \(\dfrac{-9}{4}\) < 0; \(\dfrac{1}{3}\) > o

\(\dfrac{-9}{4}\) < \(\dfrac{1}{3}\)

b; \(\dfrac{-8}{3}\) < - 1

    \(\dfrac{4}{-7}\) > - 1

Vậy  \(\dfrac{-8}{3}\) < \(\dfrac{4}{-7}\)

c; \(\dfrac{9}{-5}\) < - 1

\(\dfrac{7}{-10}\)  > - 1

Vậy \(\dfrac{9}{-5}\) < \(\dfrac{7}{-10}\)

Câu 2:

a; Viết các phân số theo thứ tự tăng dần 

\(\dfrac{-1}{2}\); \(\dfrac{2}{7}\); \(\dfrac{2}{5}\)

b;  \(\dfrac{-11}{4}\); \(\dfrac{-7}{3}\); \(\dfrac{12}{5}\) 

Bài 6:

a: \(x^2+x+1\)

\(=x^2+x+\frac14+\frac34\)

\(=\left(x+\frac12\right)^2+\frac34\ge\frac34\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

b: \(2+x-x^2\)

\(=-\left(x^2-x-2\right)\)

\(=-\left(x^2-x+\frac14-\frac94\right)=-\left(x-\frac12\right)^2+\frac94\le\frac94\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

c: \(x^2-4x+1\)

\(=x^2-4x+4-3\)

\(=\left(x-2\right)^2-3\ge-3\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

d: \(4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

e: \(3x^2-6x+1\)

\(=3\left(x^2-2x+\frac13\right)\)

\(=3\left(x^2-2x+1-\frac23\right)=3\left(x-1\right)^2-2\ge-2\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

f: \(x^2-2x+y^2-4y+6\)

\(=x^2-2x+1+y^2-4y+4+1\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi x-1=0 và y-2=0

=>x=1 và y=2

g: \(h\left(h+1\right)\left(h+2\right)\left(h+3\right)\)

\(=\left(h^2+3h\right)\left(h^2+3h+2\right)\)

\(=\left(h^2+3h+1\right)^2-1\ge-1\forall h\)

Dấu '=' xảy ra khi \(h^2+3h+1=0\)

=>\(h^2+3h+\frac94=\frac54\)

=>\(\left(h+\frac32\right)^2=\frac54\)

=>\(h+\frac32=\pm\frac{\sqrt5}{2}\)

=>\(h=-\frac32\pm\frac{\sqrt5}{2}\)

Bài 5:

a: \(a^2+2a+b^2+1\)

\(=a^2+2a+1+b^2\)

\(=\left(a+1\right)^2+b^2\ge0\forall a,b\)

b: \(x^2+y^2+2xy+4\)

\(=\left(x^2+2xy+y^2\right)+4\)

\(=\left(x+y\right)^2+4\ge4>0\forall x,y\)

c: \(\left(x-3\right)\left(x-5\right)+2\)

\(=x^2-8x+15+2\)

\(=x^2-8x+17=x^2-8x+16+1=\left(x-4\right)^2+1>0\forall x\)