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fenix001 [56]
3 years ago
7

Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrog

ens. (There are several possible structures. It is enough to draw any one of them, but you may draw two or more if you want to.)

Chemistry
1 answer:
goldfiish [28.3K]3 years ago
7 0

Answer:

The structures are shown in the figure.

Explanation:

The primary hydrogens are those which are attached to primary carbon.

Primary carbons are the carbons which are attached to only one carbon.

Primary carbons is bonded to three hydrogens.

In order to draw such structure we will draw structures which will have carbon with three hydrogens or no hydrogens (quaternary)

The structures are shown in the figure with clear marking.

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I’m pretty sure they are, but it depends on which acids. Some acids have a really high or low ph that can burn through items, like our stomach acid. Instead, something with a high melting point would be used.
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Which of the following experimental observations proved that the idea of matter and mass being evenly distributed in an atom was
Ksenya-84 [330]

A. Less than 1% of the alpha particles went un-deflected through the gold foil.

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Perform the following operation
Yanka [14]

Taking into account the scientific notation, the result of the sum is 10.84300×10³.

  • <u><em>Scientific notation</em></u>

First, remember that scientific notation is a quick way to represent a number using powers of base ten.

The numbers are written as a product:

a×10ⁿ

where:

  • a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
  • n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.

  • <u><em>Sum in scientific notation</em></u>

You want to add two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 3.

Then all the values ​​are expressed as a function of the base 10 exponent with the highest exponent. In this case: 9.7300×10²= 0.97300×10³

Taking the quantities to the same exponent, all you have to do is add what was previously called the number "a". In this case:

0.97300×10³ + 9.8700×10³= (0.97300+ 9.8700)×10³= 10.84300×10³

Finally, the result of the sum is 10.84300×10³.

Learn more:

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3 0
2 years ago
Which property of gases best explains the ability of air bags to cushion the force of impact during a car accident?
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Gases are compressible
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3 years ago
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When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M = 163 m/z) is formed.
DanielleElmas [232]

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.

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