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1 year ago

Which spectroscopic tool would be best for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane?

1 answer:

5 0

1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

The preferred method for determining or validating the structure of organic molecules or those containing protons is H NMR. When compared to other nuclei, a solution-state proton spectrum may be obtained relatively quickly, and it contains a wealth of knowledge regarding a compound's structure.

It can be calculated by simply counting the number of unique hydrogens on one side of the symmetry plane will give you the count of signals individual molecules emit in a 1H NMR spectrum.

Therefore, 1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

To know more about 1-H NMR spectroscopy

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Answer:

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3 years ago

What is the mass of 0.714 moles of Mercury (I) Chloride (Hg2Cl2)?

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Data:

m (Sample Mass) = ?
n (Number of moles) = 0.714 mol
MM (Molar Mass) of Mercury (I) Chloride ( $Hg_{2} Cl_{2}$ )
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass $Hg_{2} Cl_{2}$ = 401.18 + 70.906 = 472.086 ≈ 472.09 amu or 472.09 g/mol

Formula:

$n = \frac{m}{MM}$

Solving:

$n = \frac{m}{MM}$
$0.714 = \frac{m}{472.09}$
$m = 337.07226\:\to\:\boxed{\boxed{m\approx 337 g}} \end{array}}\qquad\quad\checkmark$

Answer:
By approximation would be letter D) 337.2 g

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Answer:

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Explanation:

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Answer: Aluminium is getting oxidized in the given chemical reaction.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2Al(s)+3FeSO_4(aq.)\rightarrow Al_2(SO_4)_3(aq.)+3Fe(s)$

The half cell reactions for the above reaction follows:

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Reduction half reaction: $Fe^{2+}+2e^-\rightarrow Fe$

As, aluminium is loosing 3 electrons to form aluminium cation. Thus, it is getting oxidized. Iron is gaining 2 electrons to form iron anion. Thus, it is getting reduced.

Hence, the oxidized species of the given reaction is aluminium.

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