Answer :
Sure, let’s go through each question and solve it step by step:
### Q1. What piece of equipment does Joshua need to measure the amount of salt in the 200 mL of sea water?
To measure the amount of salt in 200 mL of sea water, Joshua needs a salinometer or a refractometer. Both of these instruments are commonly used to measure the salinity or concentration of salt in a liquid.
### Q2. Joshua's three readings for the mass of salt after evaporation are 6.5 g, 7.0 g, and 7.5 g. What value for the weight/mass of salt should he use?
To determine the appropriate value for the weight or mass of salt, Joshua should calculate the average of the three readings.
The readings are:
- [tex]\(6.5 \, \text{g}\)[/tex]
- [tex]\(7.0 \, \text{g}\)[/tex]
- [tex]\(7.5 \, \text{g}\)[/tex]
The formula to calculate the average is:
[tex]\[ \text{Average} = \frac{\text{Sum of readings}}{\text{Number of readings}} \][/tex]
Substituting the given readings:
[tex]\[ \text{Average} = \frac{6.5 + 7.0 + 7.5}{3} = \frac{21.0}{3} = 7.0 \, \text{grams} \][/tex]
Therefore, Joshua should use an average mass of [tex]\(7.0 \, \text{grams}\)[/tex] for the salt.
### Q3. Joshua was very accurate with his measurement of the volume of sea water so what is the concentration of the seawater in his experiment? Give your answer in grams per liter.
To find the concentration of the seawater, we need to convert the volume measurement to liters and then determine the concentration in grams per liter.
The volume of seawater given is [tex]\(200 \, \text{mL}\)[/tex].
Convert this volume to liters:
[tex]\[ 200 \, \text{mL} = \frac{200}{1000} \, \text{L} = 0.2 \, \text{L} \][/tex]
The average mass of the salt dissolved in this volume is [tex]\(7.0 \, \text{grams}\)[/tex].
Concentration is defined as the mass of solute (salt) per unit volume of solution (seawater):
[tex]\[ \text{Concentration} = \frac{\text{Mass of salt}}{\text{Volume in liters}} \][/tex]
Substitute the values:
[tex]\[ \text{Concentration} = \frac{7.0 \, \text{grams}}{0.2 \, \text{liters}} = 35.0 \, \text{grams per liter} \][/tex]
Therefore, the concentration of the seawater in Joshua’s experiment is [tex]\(35.0 \, \text{grams per liter}\)[/tex].
### Q1. What piece of equipment does Joshua need to measure the amount of salt in the 200 mL of sea water?
To measure the amount of salt in 200 mL of sea water, Joshua needs a salinometer or a refractometer. Both of these instruments are commonly used to measure the salinity or concentration of salt in a liquid.
### Q2. Joshua's three readings for the mass of salt after evaporation are 6.5 g, 7.0 g, and 7.5 g. What value for the weight/mass of salt should he use?
To determine the appropriate value for the weight or mass of salt, Joshua should calculate the average of the three readings.
The readings are:
- [tex]\(6.5 \, \text{g}\)[/tex]
- [tex]\(7.0 \, \text{g}\)[/tex]
- [tex]\(7.5 \, \text{g}\)[/tex]
The formula to calculate the average is:
[tex]\[ \text{Average} = \frac{\text{Sum of readings}}{\text{Number of readings}} \][/tex]
Substituting the given readings:
[tex]\[ \text{Average} = \frac{6.5 + 7.0 + 7.5}{3} = \frac{21.0}{3} = 7.0 \, \text{grams} \][/tex]
Therefore, Joshua should use an average mass of [tex]\(7.0 \, \text{grams}\)[/tex] for the salt.
### Q3. Joshua was very accurate with his measurement of the volume of sea water so what is the concentration of the seawater in his experiment? Give your answer in grams per liter.
To find the concentration of the seawater, we need to convert the volume measurement to liters and then determine the concentration in grams per liter.
The volume of seawater given is [tex]\(200 \, \text{mL}\)[/tex].
Convert this volume to liters:
[tex]\[ 200 \, \text{mL} = \frac{200}{1000} \, \text{L} = 0.2 \, \text{L} \][/tex]
The average mass of the salt dissolved in this volume is [tex]\(7.0 \, \text{grams}\)[/tex].
Concentration is defined as the mass of solute (salt) per unit volume of solution (seawater):
[tex]\[ \text{Concentration} = \frac{\text{Mass of salt}}{\text{Volume in liters}} \][/tex]
Substitute the values:
[tex]\[ \text{Concentration} = \frac{7.0 \, \text{grams}}{0.2 \, \text{liters}} = 35.0 \, \text{grams per liter} \][/tex]
Therefore, the concentration of the seawater in Joshua’s experiment is [tex]\(35.0 \, \text{grams per liter}\)[/tex].
