CTC: MO

\(MO+H_2SO_4-->MSO_4+H_2O\)

x................x................x..................x

\(m_{ddH_2SO_4}=\dfrac{x.98.100\%}{14\%}=700x\left(g\right)\)

\(m_{dd}=x\left(M+16\right)+700x\left(g\right)\)

\(C\%MSO_4=\dfrac{x\left(M+96\right)}{x\left(M+16\right)+700x}.100\%=16,2\%\)

\(\Leftrightarrow\dfrac{M+96}{M+16+700}.100\%=16,2\%\)

\(\Rightarrow M\approx24\)

=>M là Mg

=> CTHH MgO

http://violet.vn/thcs-levanhien-tuyenquang/present/show/entry_id/8933448

bài 3 bn có thể tham khảo đây còn mik k hiểu chỗ a lắm nên k làm dc

\(A=5^{16}-\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\)

    \(=5^{16}-\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\)

   \(=5^{16}-\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\)

  \(=5^{16}-\left(5^8-1\right)\left(5^8+1\right)\)

\(=5^{16}-\left(5^{16}-1\right)=1<2005\)

tick mình đi rồi mình giải cho

\(n_{NaOH}=\dfrac{100.8\%}{40.100\%}=0,2\left(mol\right)\)

\(NaOH+HCl-->NaCl+H_2O\)

0,2..............0,2.................0,2

\(V_{HCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

\(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

a) Sắt + axit clohidric --------> sắt(II) clorua +hidro

\(Fe+2HCl-->FeCl_2+H_2\uparrow\)

b) \(m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\)

\(m_{HCl}=m_{FeCl_2}+m_{H_2}-m_{Fe}=34,28+0,54-15=19,82\left(g\right)\)

\(n_{CaCO_3}=0,05\left(mol\right)\)

\(n_{HCl}=\dfrac{100.3,65\%}{36,5.100\%}=0,1\left(mol\right)\)

\(CaCO_3+2HCl-->CaCl_2+H_2O+CO_2\uparrow\)

\(\dfrac{0,05}{1}=\dfrac{0,1}{2}\) => 2 chất hết

dd sau phản ứng CaCl2

\(C\%CaCl_2=\dfrac{0,1.36,5}{5+100-0,05.44}.100\%\approx3,55\%\)

\(\left\{{}\begin{matrix}e=p\\p+n=\dfrac{13}{7}e\\e+p+n=40\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=14\\e=14\\n=12\end{matrix}\right.\)

mạch điện đâu bạn ? hoặc cho bt mạch nt hay // chứ ?

\(P=10m=60.10=600\left(N\right)\)

\(A_i=P.h=600.5=3000\left(J\right)\)

\(A_{ms}=F_{ms}.l=25.40=1000\left(J\right)\)

\(A_{tp}=A_i+A_{ms}=3000+1000=4000\left(J\right)\)

\(H=\dfrac{A_i}{A_{tp}}.100\%=\dfrac{3000}{4000}.100\%=75\%\)