In the table below, the experimental data collected by Students A, B and C are given on lines 1, 2 and 3. Calculate the missing values for these students, and enter these values on lines 4 to 8. Be sure to use the correct number of significant figures. In the spaces marked "Set-up", show your work to calculate the values for Student A only. Complete all the calculations and then average the results of the three trials. On the assigned date, submit to your instructor the report form (page 2) and the complete calculations for the exercises on page 3. |DATA|| Student A Student B Student C Line# 1 Weight of the evaporating dish 46.075 g 51.857 g 48.432 g 2 Volume of the salt solution 5.00 mL5.00 mL5.00 mL 3 Weight of the dish and residue after evaporation 4Weight of dry sodium chloride Set-up: Line #3 – Line #1 46.587 g 52.355 g 48.937 g 5Moles of sodium chloride Set-up: Line #4÷58.5 g/mol moles moles moles 6Volume of solution in liters Set-up: #2 1000 mL
States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
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![**Problem:**
3. 5.0 mL of a potassium chloride solution is transferred to an evaporating dish and the water evaporated. The weight of KCl is 0.75 g.
a. Calculate the molarity
b. Calculate the percent (w/v) KCl
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**Solution:**
**a. Calculate the molarity**
To calculate the molarity (M), use the formula:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
1. **Find the moles of KCl**:
- The molar mass of KCl is approximately 74.55 g/mol.
- Moles of KCl = \[ \frac{0.75 \text{ g}}{74.55 \text{ g/mol}} \]
2. **Convert volume to liters**:
- Volume = 5.0 mL = 0.005 L
3. **Calculate the molarity**:
- Molarity (M) = \[ \frac{\text{moles of KCl}}{0.005 \text{ L}} \]
**b. Calculate the percent (w/v) KCl**
Percent (w/v) is calculated using the formula:
\[ \text{Percent (w/v)} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \right) \times 100 \]
1. Substituting the given values:
- Mass of KCl = 0.75 g
- Volume of solution = 5.0 mL
- Percent (w/v) = \[ \left( \frac{0.75 \text{ g}}{5.0 \text{ mL}} \right) \times 100 \]
This setup lays a foundational understanding for solving problems involving the concentration of solutions and can be applied to various chemical calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06647c36-82a9-4e7a-9695-c17c8e816387%2F21dd03b3-930d-4017-b9f5-abd843e94b85%2Fzl5dv1.jpeg&w=3840&q=75)

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