Let's solve the problem step by step:
### Part A: Length of the cube of \( \text{Na}_2\text{CS}_3 \)
Given:
- Volume of \( \text{CS}_2 \) = \( 88.0 \text{ mL} \)
- Density of \( \text{CS}_2 \) = \( 1.26 \text{ g/mL} \)
- Volume of \( \text{NaOH} \) = \( 750.0 \text{ mL} \)
- Molarity of \( \text{NaOH} \) = \( 3.76 \text{ M} \)
- Density of \( \text{Na}_2\text{CS}_3 \) = \( 2.36 \text{ g/cm}^3 \)
1. Calculate the volume of \( \text{Na}_2\text{CS}_3 \) that will be produced:
The volume is calculated to be \( 62.980 \text{ cm}^3 \).
2. Determine the side length of the cube:
The volume of a cube is given by \( V = \text{side}^3 \). Hence, we can find the side length by:
[tex]\[
\text{side length} = \sqrt[3]{\text{volume}}
\][/tex]
Substituting the volume:
[tex]\[
\text{side length} = \sqrt[3]{62.980 \text{ cm}^3}
\][/tex]
The cube's side length in mm is found to be:
[tex]\[
\text{side length} \approx 39.786 \text{ mm}
\][/tex]
### Part B: Volume of water formed
Given:
- Density of water = \( 0.997 \text{ g/cm}^3 \)
1. Calculate the volume of water produced in \( \text{cm}^3 \):
The volume of water produced is calculated to be \( 76.433 \text{ mL} \).
2. Convert this volume to liters (L):
Note that \( 1 \text{ cm}^3 = 1 \text{ mL} \) and \( 1000 \text{ mL} = 1 \text{ L} \), thus:
[tex]\[
\text{Volume in L} = \frac{76.433 \text{ mL}}{1000} = 0.076433 \text{ L}
\][/tex]
### Summary of Results:
a) The side length of the cube of \( \text{Na}_2\text{CS}_3 \) produced is approximately \( 39.786 \text{ mm} \).
b) The volume of water formed under the given conditions is approximately [tex]\( 0.076433 \text{ L} \)[/tex].
