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kolbaska11 [484]
3 years ago
5

One isotope of a metallic element has the mass number 67 and 37 neutrons in the nucleus. The cation derived from the isotope has

28 electrons. Write the atomic symbol for this element.
Chemistry
2 answers:
Anuta_ua [19.1K]3 years ago
6 0
No. Of protons = mass no. - no. Of neutrons
No. Of protons = 67 - 37
= 30
No. Of electrons = 28

Zinc will have 30 protons and 28 electrons. So, it will have +2 charge

Symbol - Zn^+2
miv72 [106K]3 years ago
5 0

Answer:

Zinc: Zn

Explanation:

The mass number of element is 67

mass number = number of protons + number of neutrons

So number of protons = 67-37=30

Thus the atomic number of element is 30

the atom with atomic number 30 is zinc

On losing 2 electrons it will form dipositive ion which will have 28 electrons

The symbol of element is Zn

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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.
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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.

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