drag the tiles to the correct boxes to complete the pairs use this table to match the molecules with their estimated masses

We want to add the indicated atoms mass for each tile
Since each hydrogen atom's mass corresponds to
[tex]1.67\cdot10^{-24}[/tex]Then H₂:
[tex]\begin{gathered} H_2=2\cdot H \\ H_2=2\cdot1.67\cdot10^{-24} \end{gathered}[/tex]we multiply just the numbers without exponent:
[tex]\begin{gathered} H_2=2\cdot1.67\cdot10^{-24} \\ =3.34\cdot10^{-24} \end{gathered}[/tex]Similarly, to the previous, we know that:
[tex]\begin{gathered} CO_2=C+2\cdot O \\ CO_2=1.99\cdot10^{-23}+2\cdot2.66\cdot10^{-23} \end{gathered}[/tex]First we multiply the numbers without exponent: 2 * 2.66:
[tex]\begin{gathered} 2\cdot2.66=5.32 \\ CO_2=1.99\cdot10^{-23}+5.32\cdot10^{-23} \end{gathered}[/tex]Now, we can add both terms:
[tex]\begin{gathered} CO_2=1.99\cdot10^{-23}+5.32\cdot10^{-23} \\ CO_2=(1.99^{}+5.32)\cdot10^{-23} \\ CO_2=7.31\cdot10^{-23} \end{gathered}[/tex]We know that:
[tex]\begin{gathered} NO_2=N+2\cdot O \\ NO_2=2.32\cdot10^{-23}+2\cdot2.66\cdot10^{-23} \end{gathered}[/tex]We can see this is really similar to the previous one.
We follow the same procedure:
First we multiply the numbers without exponent: 2 * 2.66:
[tex]\begin{gathered} 2\cdot2.66=5.32 \\ NO_2=2.32\cdot10^{-23}+5.32\cdot10^{-23} \end{gathered}[/tex]Now, we can add both terms:
[tex]\begin{gathered} NO_2=2.32\cdot10^{-23}+5.32\cdot10^{-23} \\ NO_2=(2.32^{}+5.32)\cdot10^{-23} \\ NO_2=7.64\cdot10^{-23} \end{gathered}[/tex]We know that:
[tex]\begin{gathered} NH_3=N+3\cdot H \\ NH_3_{}=2.32\cdot10^{-23}+3\cdot1.67\cdot10^{-24} \end{gathered}[/tex]First we multiply the numbers without exponent: 3 * 1.67:
[tex]\begin{gathered} 3\cdot1.67=5.01 \\ NH_3=2.32\cdot10^{-23}+5.01\cdot10^{-24} \end{gathered}[/tex]For the first time we have two terms with different 10 power:
on one side 10⁻²³ and on the other 10⁻²⁴
We want both have the same because if they are different we cannot add them.
Then we want to transform
5.01 · 10⁻²⁴
Into
___ · 10⁻²³
We need to find that number
Since
[tex]10^{-1}\cdot10^{-23}=10^{-24}[/tex]Then
[tex]5.01\cdot10^{-1}\cdot10^{-23}=5.01\cdot10^{-24}[/tex]We know that
[tex]\begin{gathered} 10^{-1}=\frac{1}{10}=0.1 \\ \downarrow \\ 5.01\cdot10^{-1}=5.01\cdot0.1=0.501 \end{gathered}[/tex]Then
[tex]5.01\cdot10^{-24}=5.01\cdot10^{-1}\cdot10^{-23}=0.501\cdot10^{-23}[/tex]Then, getting back to our equation for NH₃, we have that:
[tex]\begin{gathered} NH_3=2.32\cdot10^{-23}+5.01\cdot10^{-24} \\ NH_3=2.32\cdot10^{-23}+0.501\cdot10^{-23} \end{gathered}[/tex]Now, we can add normally as we did on the previous molecules:
[tex]\begin{gathered} NH_3=2.32\cdot10^{-23}+0.501\cdot10^{-23} \\ NH_3=(2.32+0.501)\cdot10^{-23} \\ NH_3=2.821\cdot10^{-23} \end{gathered}[/tex]