Your answer is 3.25 moles of Bromine
Answer:
The IUPAC structure only shows bond pairs and lone pairs. In the flouromethane structure above, there is only one bond pair and three lone pairs of electrons. Therefore there is one electron remaining, but since it doesn't not make up a pair, it is ignored in the structure but theoretically it is present.

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml
<h3>Subtraction of Numbers</h3>
Given Data
- Volume of Hexane dispensed = 40ml
Let us compute the amount of excess hexane/ the volume that will remain
Remainder = The difference in volume dispensed and the volume needed
Remainder = 40-5
Remainder = 35 ml
The remainder is 35ml
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Answer:
The answer is quartet 2.40 ppm.
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Explanation:
Solution
Multiplicity or (n+1) rule:
It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.
Multiplicity =(n+1)
Thus
The non equivalent protons which are attached from adjacent atom is denoted by n.
Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:
The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:
Non- equivalent = 3
Multiplicity = (3 +1)
= 4
= Quartet for 2H
A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.
Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.
The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
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