To determine the missing values in the ratios for the given compounds, let's analyze their chemical formulas carefully.
### Calcium bicarbonate ([tex]$\text{Ca(HCO}_3\text{)}_2$[/tex]):
For calcium bicarbonate, the chemical formula is [tex]$\text{Ca(HCO}_3\text{)}_2$[/tex]. This formula tells us that there is one calcium ion ([tex]$\text{Ca}^{2+}$[/tex]), and two bicarbonate ions ([tex]$\text{HCO}_3^-$[/tex]).
Each bicarbonate ion ([tex]$\text{HCO}_3^-$[/tex]) contains:
- 1 Hydrogen atom (H)
- 1 Carbon atom (C)
- 3 Oxygen atoms (O)
Since there are two bicarbonate ions, we multiply the number of atoms in one [tex]$\text{HCO}_3^-$[/tex] by 2:
- Hydrogen (H): [tex]\( 2 \times 1 = 2 \)[/tex]
- Carbon (C): [tex]\( 2 \times 1 = 2 \)[/tex]
- Oxygen (O): [tex]\( 2 \times 3 = 6 \)[/tex]
So, for the full compound ([tex]$\text{Ca(HCO}_3\text{)}_2$[/tex]), we have:
- Calcium (Ca): 1 atom
- Hydrogen (H): 2 atoms
- Carbon (C): 2 atoms
- Oxygen (O): 6 atoms
Thus, the ratio of Ca:H:C:O is:
[tex]\[ 1 : 2 : 2 : 6 \][/tex]
### Lithium sulfide ([tex]$\text{Li}_2\text{S}$[/tex]):
For lithium sulfide, the chemical formula is [tex]$\text{Li}_2\text{S}$[/tex]. This formula tells us that there are:
- 2 Lithium atoms (Li)
- 1 Sulfur atom (S)
Thus, the ratio of [tex]$\text{Li} : \text{S}$[/tex] is:
[tex]\[ 2 : 1 \][/tex]
### Conclusion:
For the given compounds, we have:
- The ratio for [tex]$\text{Ca(HCO}_3\text{)}_2$[/tex] (Calcium bicarbonate) is [tex]$\text{Ca:H:C:O} = 1:2:2:6$[/tex]
- The ratio for [tex]$\text{Li}_2\text{S}$[/tex] (Lithium sulfide) is [tex]$\text{Li:S} = 2:1$[/tex]
Therefore, filling in the missing values:
[tex]\[
\begin{array}{ccc}
\text{Calcium bicarbonate } \left( \text{Ca(HCO}_3\text{)}_2 \right) & \quad \text{Lithium sulfide } \left( \text{Li}_2\text{S} \right) \\
\text{Ca : H : C : O} & \quad \text{Li : S} \\
1 : 2 : 2 : \boxed{6} & \quad 2 : \boxed{1}
\end{array}
\][/tex]
