1.

\(a,b,c\): vật lí.

\(d\): hóa học.

2.

\(N_2O_5=\dfrac{5.II}{2}=V\)

\(BaO=\dfrac{1.II}{1}=II\)

\(HCl=\dfrac{1.I}{1}=I\)

\(H_2SO_4=\dfrac{2.I}{4}=\dfrac{II}{IV}=\dfrac{I}{II}\)

3.

a. \(CTDC:NaSO_4\)

\(Na_x(SO_4)_y=\dfrac{x}{y}=\dfrac{II}{I}=\dfrac{2}{1}\Leftrightarrow\begin{cases}x=2\\y=1\end{cases}\)

\(\Rightarrow CTHH:Na_2SO_4\)

b. \(CTDC: SiO\)

\(Si_xO_y=\dfrac{x}{y}=\dfrac{II}{IV}=\dfrac{2}{4}=\dfrac{1}{2}\Leftrightarrow\begin{cases}x=1\\y=2\end{cases}\)

\(\Rightarrow CTHH:SiO_2\)

c. \(CTDC: FeBr\)

\(Fe_xBr_y=\dfrac{x}{y}=\dfrac{I}{III}=\dfrac{1}{3}\Leftrightarrow\begin{cases}x=1\\y=3\end{cases}\)

\(\Rightarrow CTHH:FeBr_3\)

d. \(CTDC: CaN\)

\(Ca_xN_y=\dfrac{x}{y}=\dfrac{III}{II}=\dfrac{3}{2}\Leftrightarrow\begin{cases}x=3\\y=2\end{cases}\)

\(\Rightarrow CTHH:Ca_3N_2\)

4. 

- \(KClO_3\):

+\(K,Cl,O\)

+ \(K,Cl,3O\)

+ \(PTK=39+35,5+3.16=122,5đvC\)

- \(H_2SO_4\):

+ \(H,S,O\)

+ \(2H, S, 4O\)

+ \(PTK= 2.1+32+4.16=98đvC\)

- \(N_2O_5\):

+ \(N,O\)

+ \(2N,5O\)

+ \(PTK=2.14+5.16=108đvC\)

- \(CuCO_3\):

+ \(Cu,C,O\)

+ \(Cu,C,3O\)

+ \(PTK=64+12+3.16=124đvC\)

5.

\(p+e+n=26\)

\(p+e-n=22\)

\(\Rightarrow 2(p+e)=48\)

\(p+e=24\)

Mà \(p=e=\dfrac{24}{2}=12\)

\(n=26-24=2\)

Vậy \(p=12;e=12;n=2\)