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natka813 [3]
2 years ago
14

A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p

late under a UV lamp, the student found their sample had two spots corresponding to Rf values of 0.31 and 0.85. They ran the TLC plate again, this time dissolving the compound in dicholoroacetic acid and using ethyl acetate as the eluent. What will be the Rf values for each spot?
Chemistry
1 answer:
Zepler [3.9K]2 years ago
3 0

1) Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:

<span><span>Determine the Rf values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent.</span>Which compound, A, B, or C, is the most polar?<span>What would you expect to happen to the Rf values if you used acetone instead of hexanes as the eluting solvent?</span><span>How would the Rf values change if eluted with hexanes using an alumina TLC plate?</span></span>

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2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?

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3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

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4) A student spots an unknown sample on a TLC plate. After developing in hexanes/ethyl acetate 50:50, he/she saw a single spot with an Rf of 0.55. Does this indicate that the unknown material is a pure compound? What can be done to verify the purity of the sample?

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5) Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC plate and developed in a hexanes/ethyl acetate solvent mixture. Predict the relative Rf values for the three compounds in the sample.

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6) Plate A, below, represents the TLC chromatogram of a compound run in hexanes. The same compound was then spotted on a large TLC plate and again run in hexanes. Which TLC plate, B, C, or D, correctly represents how far the compound would run on the longer plate?

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Back to TLC

Original content © University of Colorado at Boulder, Department of Chemistry and Biochemistry.
The information on these pages is available for academic use without restriction.
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Give two areas where the compressible nature of gas is applied​
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1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.

2. Balloons can be easily filled with air.

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Liquid X is known to have a higher surface tension and higher viscosity than Liquid Y Use these facts to predict the result of b
creativ13 [48]

Answer:

O FX will be greater than FY

Explanation:

<em>Surface tension</em> can be defined as the force required to stretch one film of a given fase (usually with liquids).

This required force is proportional to the liquid's surface tension. This means that the higher the surface tension, the higher the required force to stretch it is.

5 0
2 years ago
what property of the noble gases most likely prevented the gases from being readily/easily discovered?
Vlad [161]

These gases very rarely react, with others and also noble gases are odourless and colourless.

Explanation:

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7 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
2 years ago
Read 2 more answers
If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?
blagie [28]

Answer:

The answer to your question is     V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

                V1/T1 = V2/T2

-Solve for V2

                V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

                V2 = (750 x 328) / 298

-Simplification

                V2 = 246000 / 298

-Result

                V2 = 825.5 ml

             

7 0
2 years ago
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