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Sladkaya [172]
3 years ago
14

You plan to separate a polar and a non-polar compound using normal phase column chromatography technique. Methylene chloride (CH

2Cl2) and hexanes (C6H14)are the solvents to be used to elute the column. Which solvent should be used first
Chemistry
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

Hexane should be used first.

Explanation:

Chromatography is a method of separating the constituents of a mixture by taking advantage of their different rates of movement in a solvent over an adsorbent medium. Chromatography is a means of separation and analysis that utilises fractional separation. It is based on the principle  that if a fluid containing a number of substances is allowed to pass though an adsorbent medium, the different substances in the fluid may travel at different rates and be separated.

The rate of movement depends on the relative affinities of the constituents for the solvent and adsorbent medium. i.e solutes which are weakly adsorbed by the adsorbent medium are easily redissolved by the ascending solvent and quickly travel up the adsorbent medium. In addition to that , solutes which are very soluble in the solvent move up at a faster rate than those which  are not soluble.

In column Chromatography;

A non-polar solvent should be initiated and applied first. This is because , in a column chromatography, a non-polar compound will be removed at first then later polar compound.

Assuming a polar compound is used first, the polar compounds will be removed alongside with all the non-polar compounds.

From the two Compounds given;

We know that :

Hexane is a non-polar compound and Methylene chloride  is a polar compound. As such, Hexane should be used first.

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marshall27 [118]

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8 0
3 years ago
What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
asambeis [7]
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~ </span>
6 0
3 years ago
Help please
Lana71 [14]

Answer:

Yes

Explanation:

Based on the graph, nuclear energy is one of the least contributors of CO2 emissions.

You would support this source of energy .

8 0
3 years ago
A 5 cm3 piece of aluminum has a higher density than a 10cm3 of aluminum. True or false
Lubov Fominskaja [6]

Density stays the same, false

3 0
2 years ago
HURRY!!!! TIMED!!! WILL GIVE BRAINLIST!!!
aev [14]

Explanation:

(a) write a balanced equation for the reaction

CaCO3 + HCl --> CaCl2 + H2O + CO2

The balanced equation is given as;

CaCO3 + 2HCl → CaCl2 + H2O + CO2  

(b) when the reaction was complete, 800 mL of carbon dioxide gas was collected. How many moles of calcium carbonate were used in the creation?

From the balanced reaction;

1 mol of CaCO3 reacts to produce 1 mol of CO2

1 mol of CO2 = 22.4 L of CO2

This means;

1 mol of CaCO3 reacts to produce 22.4L  of CO2

x mol would produce 800ml (0.8 L) of CO2

1 = 22.4

x = 0.8

x = 0.8 * 1 / 22.4 = 0.0357 mol

(c) How many grams of CaCO3 were used?

Mass = Number of moles * Molar mass

Molar mass of CaCO3 = 100.0869 g/mol

Mass = 0.0357 mol * 100.0869 g/mol = 3.57 g

(d) If there was another contaminant in the sample that was not un reactive, would this have caused the percent yield of carbon dioxide to be higher, lower, or the same, explain your answer.

The same

An un reactive contaminant in the sample is most likely a catalyst. Catalysts only affect the rate of reaction. They do not affect yields of products.

7 0
2 years ago
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