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Lemur [1.5K]
3 years ago
10

On occasion, it has been found that the oxidation of borneol doesn't go to completion (possibly because of poor stirring or insu

fficient Oxone). This reaction would probably be easy to monitor via TLC, however. Which component should have a lower Rf, and why
Chemistry
1 answer:
seraphim [82]3 years ago
3 0

Answer:

Check the explanation

Explanation:

Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.

To answer your question:

N is oxidized from an oxidation number of -3 to an oxidation number of -1.

Cl is reduced from oxidation number of +1 to an oxidation number of -1.

Now,

Borneol should have a lower Rf because of boiling point.

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10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and
Shalnov [3]

Answer:

The density is 3g/L

Explanation:

The reaction that occurs in the vessel is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

10,0g of NH₃ are:

10,0g * \frac{1mol}{17,031g} = 0,587 moles

6,50 g of O₂ are:

6,50g * \frac{1mol}{32g} = 0,203 moles

For a complete reaction of O₂ there are necessaries:

0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}

O₂ is limiting reactant. The excess moles of NH₃ are:

0,587 - 0,163 = <em>0,424 moles of NH₃</em>

These moles are:

0,424mol * \frac{17,031g}{1mol} = <em>7,22g of NH₃</em>

Knowing O₂ is limiting reactant, mass of NO and H₂O are:

0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} = <em>4,87g of NO</em>

0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} = <em>4,39g of H₂O</em>

The total mass is: 7,22g + 4,87g + 4,39g = 16,48g ≡ <em>16,5g </em>

<em>-</em><em>The same mass add in the first. By matter conservation law-</em>

As vessel volume is 5,50L, density is:

16,5g/5,50L = <em>3g/L</em>

I hope it helps!

7 0
4 years ago
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