A blue colored ice cube floating on hot water quickly melts to release its color. The cool, blue water moves downward. This down
1 answer:
You might be interested in
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.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ of
) =-1275.0
enthalpy of combustion of oxygen(Δ of
) = zero
enthalpy of combustion of carbon dioxide(Δ of
) = -393.5
enthalpy of combustion of water(Δ of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>