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  1. Print the notecards
  2. Fold each page in half along the solid vertical line
  3. Cut out the notecards by cutting along each horizontal dotted line
  4. Optional: Glue, tape or staple the ends of each notecard together
  1. Verify Front of pages is selected for Viewing and print the front of the notecards
  2. Select Back of pages for Viewing and print the back of the notecards
    NOTE: Since the back of the pages are printed in reverse order (last page is printed first), keep the pages in the same order as they were after Step 1. Also, be sure to feed the pages in the same direction as you did in Step 1.
  3. Cut out the notecards by cutting along each horizontal and vertical dotted line
To print: Ctrl+PPrint as a list

10 notecards = 3 pages (4 cards per page)

Viewing:

Postlab for Thin Layer Chromatography SD

front 1

Image: TLC plate with point of application on left with 4 spots. The ruler underneath shows a distance of 6.20 cm from point of application to solvent front and 0.50 cm to spot 1, 1.50 cm to spot 2, 4.00 cm to spot 3 and 5.50 cm to spot 4. End of image.

The Rf of substance "3" is:

  • 1
  • 0.27
  • 0
  • 0.86
  • 0.07
  • 0.63

back 1

Rf of substance 3 = 4.00 cm / 6.20 cm = 0.65

0.63

front 2

Image: TLC plate with point of application on left with 4 spots. The ruler underneath shows a distance of 6.20 cm from point of application to solvent front and 0.50 cm to spot 1, 1.50 cm to spot 2, 4.00 cm to spot 3 and 5.50 cm to spot 4. End of image.

The Rf of substance "4" is:

  • 0
  • 0.27
  • 0.86
  • 0.63
  • 1
  • 0.07

back 2

Rf of substance 4 = 5.50 cm / 6.20 cm = 0.89

0.86

front 3

The solvent moves 3 cm in about 10 minutes. Why shouldn't the experiment be stopped at that time instead of waiting 75 minutes for the solvent to move 10 cm?

back 3

More time allows for better separation of components in a mixture.

front 4

In this experiment it takes about 10 microliters of solution to produce a spot 1 cm in diameter. If the C o (NO3)2 solution contains about 6 g C o2+ per liter, how many micrograms of C o2+ ion are there in one spot?

1 microliter = 1E-6 L

1 microgram = 1E-6 g

back 4

(6 g Co2+ / 1 L) x ( 1E-6 g / 1 microgram) x (1L / 1E-6 L) = 6

WRONG

Ex.

Followed this -> https://www.physicsforums.com/threads/simple-dimensional-analysis-problem-thanks-for-any-help.954241/

Real answer:

C= m / V -> m = C x V

m = C x V

6 g Co2+ / 1 L x 10 x 10-6 g =

60 x 10-6 g = 60 micrograms

front 5

Instructor use only. 5% = 2.5

back 5

no data

front 6

Image: TLC plate with point of application on left with 4 spots. The ruler underneath shows a distance of 6.20 cm from point of application to solvent front and 0.50 cm to spot 1, 1.50 cm to spot 2, 4.00 cm to spot 3 and 5.50 cm to spot 4. End of image.

The Rf of substance "3" is:

  • 0.07
  • 0.86
  • 0.27
  • 1
  • 0.63
  • 0

back 6

Rf of substance 3 = 4.00 cm / 6.20 cm = 0.65

0.63

front 7

Image: TLC plate with point of application on left with 4 spots. The ruler underneath shows a distance of 6.20 cm from point of application to solvent front and 0.50 cm to spot 1, 1.50 cm to spot 2, 4.00 cm to spot 3 and 5.50 cm to spot 4. End of image.

The Rf of substance "4" is:

  • 0.63
  • 0.27
  • 0.07
  • 1
  • 0
  • 0.86

back 7

Rf of substance 4 = 5.50 cm / 6.20 cm = 0.89

0.86

front 8

The solvent moves 3 cm in about 10 minutes. Why shouldn't the experiment be stopped at that time instead of waiting 75 minutes for the solvent to move 10 cm?

  • More time allows for better separation of components in a mixture.
  • Instructors don't like students to be finishing their labs too quickly.
  • More time decreases the concentration of mixture components.
  • More time increases the concentration of mixture components.

back 8

More time allows for better separation of components in a mixture.

front 9

In this experiment it takes about 10 microliters of solution to produce a spot 1 cm in diameter. If the C o (NO3)2 solution contains about 6 g C o2+ per liter, how many micrograms of C o2+ ion are there in one spot?

1 microliter = 1E-6 L

1 microgram = 1E-6 g

back 9

C (concentration) = m (mass) / V (volume) -> m = C x V

m = C x V

(6 g Co2+ / 1 L) x 10 x 10-6 g =

60 x 10-6 g = 60 micrograms <- ?

front 10

Instructor use only. 5% = 2.5

back 10

no data