Explanation:
A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.
Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.
Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.
And, you should ensure that no one else is using an open flame near your experiment.
Thus, we can conclude that following statements are correct:
- You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
- You should ensure that no one else is using an open flame near your experiment.
Answer:

Explanation:
Assume you are using 1 L of water.
Then you are washing 4 L of salty oil.
1. Calculate the mass of the salty oil
Assume the oil has a density of 0.86 g/mL.

2. Calculate the mass of salt in the salty oil

3. Calculate the mass of salt in the spent water

4. Mass of salt remaining in washed oil
Mass = 172 g - 150 g = 22 g
5. Concentration of salt in washed oil

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2
2135 cm3 divided by 127 degrees celcius = x divided by 206
answer: 3460 cm3
Answer: Molarity of
in the original sample was 1.96M
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.


Now put all the given values in the formula of molarity, we get


Thus molarity of
in the original sample was 1.96M
A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.
<u>Explanation</u>:
- When dealing with dilution we will use the following equation:
M1 V1 = M2 V2
where,
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
- By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get
M1 V1 = M2 V2
20.0 mL
5.00 M = M2
500.0 mL
M2 = (20.0 mL
5.00 M) / 500.0 mL
M2 = 0.200 M.
Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.