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GenaCL600 [577]
2 years ago
6

HELP PlZ HURRY !!!!!!

Chemistry
1 answer:
kodGreya [7K]2 years ago
6 0

Answer:

\boxed {\tt d\approx 2.7 \ g/cm^3}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the marble slab is 4.24 kilograms.

The density is 1564 cubic centimeters.

Since we want to find the density, in g/cm^3, we must convert kilograms to grams. There are 1000 grams in 1 kilogram, so multiply 4.24 by 1000.

  • 1000 g= 1 kg
  • 4.24*1000=4240 grams

Therefore,

m= 4240 \  g \\v= 1564 \ cm^3

Substitute the values into the formula.

d=\frac{4240 \ g}{1564 \ cm^3}

Divide.

d= 2.71099744 \ g/cm^3

Round to the nearest tenth. The 1 in the hundredth place tells us to leave the tenth place as is.

d \approx 2.7 \ g/cm^3

The density is about 2.7 grams per cubic centimeter.

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            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

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                = 13 × 12 (here, 12 specifies number of carbon atoms)

                = 156

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                = 163 - 156

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                = 7 + 12

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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

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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.

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