Provide 3 examples of substances that are mixtures and list the substances in them
2 answers:
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Answer:
As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.
Rule of Thirteen:
First divide the parent peak value by 13 as,
= 163 ÷ 13
= 12.53
Now, multiply 13 by 12,
= 13 × 12 (here, 12 specifies number of carbon atoms)
= 156
Now subtract 156 from 163,
= 163 - 156
= 7
Add 7 into 12,
= 7 + 12
= 19 (hydrogen atoms)
So, the rough formula we have is,
C₁₂H₁₉
Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,
C₁₂H₁₉ -------N--------> C₁₁H₁₇N
Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.
C₁₁H₁₇N -------O--------> C₁₀H₁₃NO
Molecular Formula: C₁₀H₁₃NO
Also,
In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.
The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.
It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.
The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.
Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL