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Papessa [141]
3 years ago
11

Which of the following pairs of compounds would be most easily separated by thin layer chromatography: n-octyl alcohol and 1-oct

ene, hexadecane and octadecane, or chlorobenzene and bromobenzene?
A. hexadecane and octadecane
B. n-octyl alcohol and 1-octene
C. chlorobenzene and bromobenzene
Chemistry
1 answer:
vivado [14]3 years ago
8 0

Answer:

B. n-octyl alcohol and 1-octene

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. The principle is that different compounds in the sample mixture travel at different rates due to the differences in interactions with stationary phase and due to the differences in solubility in the solvent. The principal chemical property for separation using this technique is molecular polarity

You can intuit than hexadecane and octadecane don't have big polarity differences, also chlorobenzene and bromobenzene haven't.

An alcohol as n-octyl alcohol has different polarity than an alkene as 1-octene.

Thus, using  thin layer chromatography is most easy to separate:

<em>B. n-octyl alcohol and 1-octene </em>

<em></em>

I hope it helps!

<em></em>

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1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

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The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

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<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

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<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

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