Ch = sh = why and find the stuff
For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
The first thing you do before performing anything in the laboratory is to read the procedure and prepare the materials needed. Next, if you already have the solution where you are supposed to take your 20 mL sample, then have it near you. Then, prepare a volumetric flask (750 mL) and a 20-mL pipette. Wash the pipette 3 times with the sample solution. If your diluent is water, wash the flask 3 times with water. Now, get 20 mL of sample from your parent solution, then add it to the flask (previously washed with water). Finally, add water until the mark in the flask and make sure that the water added is up to the mark based on the lower meniscus reading to be accurate in the amount inside the flask. <span />
Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
molarity of HCl=0.02117*10/1000
=2.117M
Answer:
5.00 g of solute will remain undissolved at the bottom of the container
Explanation:
From the question, the solubility of the solute in the given solvent is 45.0 grams of solute per 500 grams of solvent.
Now if i pour 50.0 grams of solute into 800 grams of solvent, it means that only 45 g will dissolve in 500 g of solvent leaving the additional 5 g undissolved.
Hence, 5 g of solute will remain undissolved at the bottom of the container.
