Question

identify the number of atoms found in each chemical below:
2 h₂so₄
4 pb(no₃)₂

hydrogen atom(s) ? lead atom(s)
? sulfur atom(s) ? nitrogen atom(s)
? oxygen atom(s) ? oxygen atom(s)

Explanation:

Step1: Analyze \(2H_2SO_4\)

• For hydrogen (\(H\)): The sub - script of \(H\) is \(2\), and there are \(2\) formula units. So the number of \(H\) atoms is \(2\times2 = 4\).
• For sulfur (\(S\)): The sub - script of \(S\) is \(1\), and there are \(2\) formula units. So the number of \(S\) atoms is \(2\times1=2\).
• For oxygen (\(O\)): The sub - script of \(O\) is \(4\), and there are \(2\) formula units. So the number of \(O\) atoms is \(2\times4 = 8\).

Step2: Analyze \(4Pb(NO_3)_2\)

• For lead (\(Pb\)): The sub - script of \(Pb\) is \(1\), and there are \(4\) formula units. So the number of \(Pb\) atoms is \(4\times1 = 4\).
• For nitrogen (\(N\)): Inside the parentheses, the sub - script of \(N\) is \(1\), and there are \(2\) nitrate groups (\(NO_3\)) per formula unit and \(4\) formula units. So the number of \(N\) atoms is \(4\times2\times1=8\).
• For oxygen (\(O\)): Inside the parentheses, the sub - script of \(O\) is \(3\), and there are \(2\) nitrate groups (\(NO_3\)) per formula unit and \(4\) formula units. So the number of \(O\) atoms is \(4\times2\times3=24\).

Answer:

• For \(2H_2SO_4\): \(4\) hydrogen atoms, \(2\) sulfur atoms, \(8\) oxygen atoms.
• For \(4Pb(NO_3)_2\): \(4\) lead atoms, \(8\) nitrogen atoms, \(24\) oxygen atoms.