Sure! Let's go through the steps to find out how many moles of nickel(II) chloride [tex]$(NiCl_2)$[/tex] are added to the reaction flask:
1. Convert the volume of the solution from milliliters to liters:
The volume of the solution is given as [tex]\( 160.0 \)[/tex] milliliters (mL). To convert this to liters (L), we use the conversion factor:
[tex]\[
1 \text{ liter} = 1000 \text{ milliliters}
\][/tex]
Thus,
[tex]\[
\text{Volume in liters} = \frac{160.0 \text{ mL}}{1000} = 0.16 \text{ L}
\][/tex]
2. Calculate the number of moles of nickel(II) chloride [tex]$(NiCl_2)$[/tex]:
The concentration of the solution is given as [tex]\( 2.6 \)[/tex] mol/L. We can use the formula:
[tex]\[
\text{Number of moles} = \text{Concentration} \times \text{Volume}
\][/tex]
Substituting the given values:
[tex]\[
\text{Number of moles of } NiCl_2 = 2.6 \text{ mol/L} \times 0.16 \text{ L} = 0.416 \text{ mol}
\][/tex]
3. Round the result to 2 significant digits:
The number of moles of nickel(II) chloride calculated is [tex]\( 0.416 \)[/tex]. When rounding this to 2 significant digits:
[tex]\[
0.416 \approx 0.42 \text{ mol}
\][/tex]
To summarize, the chemist added:
- Volume in liters: [tex]\( 0.16 \)[/tex] L
- Number of moles of [tex]\( NiCl_2 \)[/tex]: [tex]\( 0.416 \)[/tex] mol
- Rounded number of moles of [tex]\( NiCl_2 \)[/tex]: [tex]\( 0.42 \)[/tex] mol
So, the chemist added approximately [tex]\( 0.42 \)[/tex] moles of nickel(II) chloride to the reaction flask.
