Answer:
it's A.
Explanation:
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(11) Deduce the number of dyes in food colouring H.
(iii) Suggest why food colouring F does not move during the experiment.
(iv) Explain which two food colourings contain the dye that is likely to be the most soluble the solvent.
(b) Determine which food colouring contains a dye with R, value closest to 0.67
Show your working.
Answer:
(ii) 1 dye
(iii) Food coloring F is insoluble in the solvent
(iv) 'E' and 'H'
(b) Food colouring G
Explanation:
Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper
The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper
(ii) The number of components into which the food colouring 'H' separates into = 1
Therefore, the number of dyes in food colouring 'H' = 1 dye
(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase
(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;
Food coloring 'E' and 'H'
(b) Using a similar question solution found on 'tutor my self' website, we have;
The [tex]R_f[/tex] values are given as follows;
[tex]R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}[/tex]
The distance moved by the solvent = 5 units
The distance moved by dyes in food colouring 'E' and 'H' = 4 units
The distance moved by dye in food colouring 'G' = 3.3 units
The distance moved by the second dye in food colouring 'E' = 2.7 units
By inspection, we get;
[tex]R_f[/tex] dye in food colouring 'G' = 3.3/5 = 0.66,
Therefore, the dye with [tex]R_f[/tex] value closest to 0.67 is the dye in food colouring 'G'.
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5). At what temperature (K) will 0.854 moles of neon gas occupy 12.3 L at 1.95
atmospheres?
Answer:
338.38 K
Explanation:
Applying,
PV = nRT............... Equation 1
Where P = pressure, V = Volume, R = Temperature, n = number of moles, T = temperature.
Make T the subject of the equation,
T = PV/nR............. Equation 2
From the question,
Given: P = 1.95 atm, V = 12.3 L, n = 0.854 moles
Constant: R = 0.083 L.atm/K.mol
Substitute these values into equation 2
T = (1.95×12.3)/(0.854×0.083)
T = 338.38 K