Answer:
In \(7,4.10^{23}\) molecules of \(AgNO_3\) are 209 g
In \(7,5.10^{23}\) molecules of \(H_2SO_4\) are 123 g
In \(9,4.10^{25}\) molecules of \(H_2\) are 312 g
Explanation:
First remember that 1 mole of any element has \(6.02 10^{23}\) molecules (this number is a constant).
we know that the molecular mass of an element relates the grams of that substance in 1 mol
To calculate how many grams of \(AgNO_3\) are in \(7,4.10^{23}\) molecules we use a rule of three
\(6,02.10^{23} molecules \longrightarrow 1mol AgNO_3\\7,4.10^{23} molecules\longrightarrow x\\x=\frac{7.,4.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23}molecules}\\x=1.23 mol AgNO_3\)
The molecular mass of \(AgNO_3\) is 169.87 g/mol
\(1 mol \longrightarrow 169.87 g\\1.23 mol\longrightarrow x\\x=\frac{1.23 mol. 169.87g}{1 mol} \\x= 209 g AgNO_3\)
In \(7,4.10^{23}\) molecules of \(AgNO_3\) are 209 g
To calculate how many grams of \(H_2SO_4\) are in \(7,5.10^{23}\) molecules we repeat the previous procedure
\(6,02.10^{23} molecules \longrightarrow 1mol H_2SO_4\\7,5.10^{23} molecules\longrightarrow x\\x=\frac{7.,5.10^{23}molecules.1mol AgNO_3 }{6,02.10^{23} molecules}\\x=1.25 mol H_2SO_4\)
The molecular mass of \(H_2SO_4\) is 98.079 g/mol
\(1 mol \longrightarrow 98.079g\\1.25 mol\longrightarrow x\\x=\frac{1.25mol. 98.079g}{1 mol}\\x= 123 g AgNO_3\)
In \(7,5.10^{23}\) molecules of \(H_2SO_4\) are 123 g
To calculate how many grams of \(H_2\) are in \(9,4.10^{25}\) molecules we repeat the previous procedure
\(6,02.10^{23} molecules \longrightarrow 1mol H_2\\9,4.10^{25} molecules\longrightarrow x\\x=\frac{9,4.10^{25}molecules.1mol H_2}{6,02.10^{23} molecules}\\x=156 mol H_2\)
The molecular mass of \(H_2\) is 2 g/mol
\(1 mol \longrightarrow 2g\\156 mol\longrightarrow x \\x=\frac{1.25mol. 98.079g}{1 mol} \\x= 312g\)
In \(9,4.10^{25}\) molecules of \(H_2\) are 312 g
