Answer:
d. supersaturated.
Explanation:
A solution naturally contains a solute and a solvent. The solute is the solid substance that dissolves in the solvent, which is usually a liquid substance. A solution has a maximum amount of solute that can dissolve in its constituent solvent.
However, when the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in the solution at that temperature, the solution is said to be SUPERSATURATED. This means that the solution contains more than the maximum amount of solute.
1) Silicon dioxide formula: SiO2 ....... 2 is a subscript for the O atom
2) From the formula you have 1 molecula of SiO2 contains 1 atom of SiO2
3) Then, 0.100 mol of SiO2 contains 0.1 mol of Si
4) Multiply by Avogadro's number: 0.100mol * 6.022*10^23 atoms/mol= 6.02*10^22 atoms
Answer: 6.02*10^22 atoms
Answer:
Molarity = 0.7 M
Explanation:
Given data:
Volume of KCl = 20 mL ( 0.02 L)
Molarity = 3.5 M
Final volume = 100 mL (0.1 L)
Molarity in 100 mL = ?
Solution:
Molarity = number of moles of solute / volume in litter.
First of all we will determine the number of moles of KCl available.
Number of moles = molarity × volume in litter
Number of moles = 3.5 M × 0.02 L
Number of moles = 0.07 mol
Molarity in 100 mL.
Molarity = number of moles / volume in litter
Molarity = 0.07 mol /0.1 L
Molarity = 0.7 M
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.
