Answer:
No, it is not appropriate to mix water and DMSO
Explanation:
We have to realize that DMSO is a highly polar solvent and water is a highly polar solvent. The question explicitly says that our target is to produce a solvent of<u><em> intermediate polarity</em></u>.
We can only do this by mixing a polar and a nonpolar solvent. We have been given the example of the mixture of acetone/hexane which is quite a perfect mixture.
Thus, it is inappropriate to mix DMSO and water.
Answer:
The amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
Explanation:
The chemical reaction between an acid and a base to form salt and water only is called a Neutralization reaction. Chemically
H⁺ + OH⁻ = H₂0
Hence, one mole of hydroxonium ion (H⁺) will combine with one mole of hydroxide ion (OH⁻) to give salt and water only.
In a completely neutralized reaction, the resulting salt is formed when there is complete dissociation of the acid and base to give salt and water with a pH of 7.
In the given question, the stated pH of between 8-9 tells us that the salt produced in this particular neutralization reaction is basic or alkaline. This usually occurs when a strong base reacts with a weak acid, producing a higher concentration of the hydroxide ion at equilibrium.
Hence the amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
If the concentration or molarity of the acid is known, then the exact amount of base required to neutralize it can be calculated. This is usually done via titrating the acid against drop wise solution of the base. Neutralization usually occurs when there is a change in colour of the resulting solution. The pH of the resulting solution can be determined using a litmus paper.
A blue litmus paper is indicative of a basic solution while a red litmus paper is indicative of an acidic solution.
To find them you would have numbers of the elements in percentage or grams then you divide them by their molar mass to get their moles. From there you divide by the smallest number. Round it to two or one sig fig. If you have a number that is for ex. 2.5 you multiply it by 2 to make it whole as well the other whole numbers. Then to find the molecular formula the problem must give you another molar mass and using your empirical formula convert it to its molar mass then you divide them, larger number over smaller number. You should get a number round it to 1 sig fig. Now you use that number and multiply the subscripts on the empirical formula to get the molecular formula.
