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2 years ago

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

1 answer:

12345 [234]2 years ago

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.

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Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(b) Hg₂²⁺(aq)

Olegator [25]

The given reaction is not spontaneous.

We must recognize changes in oxidation states that take place across elements in order to balance these equations. To accomplish this, keep in mind following guidelines:

A neutral element on its own has an oxidation number of zero.For a neutral molecule, the total number of oxidations must be zero.The net charge of an ion is equal to the sum of its oxidation numbers.In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.In a compound with no oxygen present the other halogens will also prefer -1.

One of the mercury atoms is oxidized from +1 to +2 in the simple aqueous ion, for a loss of 1 electron.

Oxidation half-reaction:

$0.5 Hg^{2+} _{2} (aq)$ → $Hg^{2+} (aq) + 1e^-$

$E^o _{ox} = - 0.92 V$

The other mercury is reduced from +1 to zero in mercury metal, for a gain of 1 electron.

Reduction half-reaction:

$0.5 Hg^{2+} _{2} (aq) + 1 e^-$ → $Hg(l)$

$E^o _{red} = 0.85V$

This is a disproportionation redox reaction !

Net reaction:

$Hg^{2+} _{2} (aq)$ → $Hg^{2+} (aq) + Hg (l)$

$E^o _{cell} = 0.85 - 0.92 = -0.07V$

The cell potential is negative so this reaction is NOT spontaneous.

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5 0

2 years ago

How do all stars form ?

professor190 [17]

Answer:

starts form from a accumulation of gas and dust ,which collapses due to gravity and starts to form stars

6 0

3 years ago

A beaker containing 80 grams of lead(ii) nitrate, pb(no3)2, in 100 grams of water has a temperature of 30 ºc. approximately how

sukhopar [10]

Answer:

14 g.

Explanation:

From the figure attached:

<em>the solubility of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC is </em><em>(66 g).</em>

When beaker containing 80 grams of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC.

<em>∴ The grams of the salt are undissolved, on the bottom of the beaker are </em><em>(14 g).</em>

4 0

4 years ago

Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 7.60 feet deep, 100.0 yards wi

Effectus [21]

Answer:

[MCHM] = 7.52 M

Explanation:

This all about unit conversion

1 feet = 0.3048 meters

1 yard = 0.9144 meters

7.60 feet . 0.3048 = 2.31 meters deep

100 yd . 0.9144 meters = 91.44 meters long and 91.44 meters wide

In the river we have a volume of:

2.31 m . 91.44m . 91.44m = 19314.5 m³

1 dm³ = 1 L

1 dm³ = 0.001 m³

19314.5 m³ / 0.001 m³ = 19314542 L (The final volume of the river)

3.785 L = 1 gallon

In 19314542 L, we have (19314542 / 3.785) = 5102917 gallons

1 gallon = 128 oz

5102917 gallons . 128 = 653173376 oz

1 oz = 28.3495 g

653173376 oz . 28.495 = 1.86x10¹⁰ grams

Molar mass MCHM = 128 g/m

Moles of MCHM = 1.86x10¹⁰ grams / 128 g/m = 145312500 moles

Molarity = mol /L → 145312500 moles / 19314542 L = 7.52 M

7 0

4 years ago

On number 5 I am really stuck can I have help please.

zmey [24]

Answer to this question is absorbed

5 0

3 years ago