\(m_{\text{CaCO_3}}=1000.95\%=950kg\\ \rightarrow n_{\text{CaCO_3}}=9,5mol\)

\(m_{CaCO_3}\underrightarrow{t^o}CaO+CO_2\)

9,5            →   9,5

\(\rightarrow V_{CO_2}=9,5.22,4=212,8\)

→  hiệu suất phản ứng là

\(\dfrac{159,6}{212,8}.100=75\%\)

b)\(CTPTcủa-x-là\left(C_6H_{10}O_5\right)_{n\cdot}\)

sơ đồ phản úng \(\left(C_6H_{10}O_5\right)_{n\rightarrow}\)

\(nC_6H_{12}O_6\rightarrow2nAg\)

\(n_{\left(C_6H_{12}O_6\right)_n}=\dfrac{16,2}{162n}=\dfrac{0,1}{n}\left(mol\right)\)

với hiệu suất quá trình bằng 80% thì

\(m_{Ag}4n.\dfrac{0,1}{n}.108.80\%=17,28gam\)

a)\(n_{CO_2}=0,6mol;n_{H_2O}=0,5mol\Rightarrow n_{H\left(X\right)}=2n_{H_2O}=0,1mol\)

ta có:\(m=m_C+m_H+m_O\Leftrightarrow16,2=0,6.12+1.1+m_O=8g\Rightarrow n_{O\left(X\right)}=0,5mol\)

\(n_C:n_H:n_O=0,6:1:0,5=6:10:5\)

\(\Rightarrow CTĐGN\) của X là \(C_6H_{10}O_5\Rightarrow X\in polisaccarit\)

\(m_{HCl}=9,55-5,9=3,65\\ \Rightarrow M_X=59\\ \Rightarrow X:C_3H_9N\)

số CT cất tạo ứng với công thức phân tử của X là:\(X:2^{3-1}=4\)

⇒chọn B

\(-\dfrac{6}{x}=\dfrac{1}{2}\\ \Rightarrow\left(-6\right)2=x\\ \Rightarrow x=-12\)

1.theo bất đẳng thức côsi ta có

\(a+b\ge2\sqrt{ab}\\ b+c\ge2\sqrt{ab}\\ c+a\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab.bc.ca}\)

                                       \(\ge8\sqrt{a^2b^2c^2}\\ \ge8abc\)

2.\(a^4+b^2\ge2\sqrt{a^4b^2}=2a^4b^2\)

\(\dfrac{a}{a^4+b^2}\le\dfrac{a}{2a^2b}=\dfrac{1}{2ab}\)

tương tự:\(\dfrac{b}{b^4+a^2}\le\dfrac{1}{2ab}\)

\(\rightarrow\dfrac{a}{a^4+b^2}+\dfrac{b}{b^4+a^2}\le\dfrac{1}{ab}\)

dấu = xảy ra khi \(a^4=b^2\\ b^4=a^2\)\(\rightarrow a^2=b^2=1\)

a)vì 20.45=30.30

b)vì -25.77=-55.35

hợp chất 1:

trong 1g hợp chất thì \(m_{CaO}=0,737g\)

                                   \(m_{SiO2}=0,263g\)

\(\rightarrow m_{CaO}=\dfrac{0,737}{56}=0,01316\left(mol\right)\)

\(n_{SiO2}=\dfrac{0,263}{60}=0,00438\left(mol\right)\)

\(\rightarrow cứ\dfrac{0,0136}{0,00438}=3molCaO\) kết hợp với \(\Delta molSiO2\)

-họp chất 2:

trong 1 g hợp chất có :\(m_{CaO}=0,651g\\ m_{SiO2}=0,349g\)

\(\rightarrow n_{CaO}=\dfrac{0,651}{56}=0,011625mol\)

\(n_{SiO2}=\dfrac{0,349}{60}=0,005817mol\)

\(\rightarrow cứ\dfrac{0,011625}{0,005817}=2molCaO\) kết hợp với \(\Delta molSiO2\)

1.\(p=\dfrac{x^2}{x-2}.\left(\dfrac{x^2+4}{x}-4\right)+3\)

\(=\dfrac{x^2}{x-2}.\dfrac{x^2-4x+4}{x}+3\)

\(=\dfrac{x^2}{x-2}.\dfrac{\left(x-2\right)^2}{x}+3\)

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

2.\(P=\left(x-1\right)^2+2\\ Pmin=2\Leftrightarrow x=1\)

a)\(\left(\dfrac{1}{x-2}-\dfrac{2x}{4-2x}+\dfrac{1}{2+x}\right)\left(\dfrac{2}{x}-1\right)\)

\(=\left[\dfrac{-\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}-\dfrac{2x}{4-2x}+\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right].\left(\dfrac{2-x}{x}\right)\)

\(=\left[\dfrac{-\left(x+2\right)-2x+2-x}{4\left(x+2\right)\left(2-x\right)}\right].\dfrac{2-x}{x}\)

\(=\dfrac{-4x}{\left(x+2\right)\left(2-x\right)}.\dfrac{2-x}{x}=-\dfrac{4}{x+2}\)

b)\(2x^2+x=0.x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{1}{2}\Rightarrow A=-\dfrac{8}{3}\end{matrix}\right.\)

c)\(A=\dfrac{1}{2}\Rightarrow-\dfrac{4}{x+2}=\dfrac{1}{2}\Leftrightarrow-8=x+2\)

                                             \(\Leftrightarrow x=-10\)

d)\(A\in Z^+\Leftrightarrow\left\{{}\begin{matrix}x+2\inư\left(-4\right)\\x+2< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2=-1\\x+2=-4\\x+2=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-6\\x=-4\end{matrix}\right.\)