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elixir [45]
2 years ago
5

New research shows that a red food colouring with an R value of 0.62 (when a particular solvent and temperature are used) could

be dangerous to certain people. Samples of food colouring from two different red sweets were collected and the chromatogram bolow was produced. For each of the sweets, state and explain whether or not you would recommend that the sweets are taken off the market based on this research, ​

Chemistry
1 answer:
svlad2 [7]2 years ago
3 0

It is recommended only for the second type of candy to be taken off the market.

Based on the background information, if a candy shows a 0,62 Rf value in the chromatography it is considered potentially dangerous, and therefore it should be taken off the market.

The Rf value is calculated using the following formula:

  • Rf =  Total distance traveled by a component/ Total distance from the pencil line to the solvent front

Now, let's find out if any of the components has an Rf equal to 0,62.

Note: To determine the distances I measured the image using a ruler and you can do the same, this will not alter the results.

  1. Sample 1
  • Total distance traveled by the component: 6cm
  • Total distance: 7 cm
  • Rf : 6 cm/ 7 cm = 0.85

     2. Sample 2

  • Total distance traveled by the component: 4.34cm
  • Total distance: 7 cm
  • Rf : 4.34 cm/ 7 cm = 0.62

This means sample 2 has the 0.62 Rf value and therefore it needs to be taken off the market.

Learn more about chromatography in: brainly.com/question/10296715

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determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)
sasho [114]

The  mass  of water formed  is


<u><em>calculation</em></u>

Use  the  ideal   gas  equation   to  calculate the  moles of  NH3  and O2

that  is  Pv= n RT

where;  P= pressure,  

V=  volume,

n = number  of  moles,

R=gas   constant  = 0.0821  l .atm/ mol.K

make n the formula of  the subject  by diving   both side  by  RT

n =  PV /RT

The   moles of NH3

n= (1.50 atm  x 12.5 L) /(  0.0821 L. atm /mol.k   x 298 K)  =0.766  moles

The  moles  of  O2

=(1.1 atm  x 18.9  L) /  (  0.0821 L. atm/ mol.k   x 323 K) = 0.784  moles


write the reaction  between  NH3  and  O2

4 NH3  + 5 O2  →4 No  +6H2O


from  equation above  0.766  moles of NH3  reacted to produce  

0.766 x 6/4 =1.149 moles of H2O


0.784  moles of O2   reacted to  produce  0.784  x 6/5=0.9408  moles  of H20


since  O2  is totally  consumed, O2  is the limiting  reagent  and therefore  the  moles of H2O  produced=  0.9408  moles


mass  of  H2O  = moles x molar mass

 from  periodic table the  molar mass  of H2O  =  (1 x2)+16= 18  g/mol

mass = 18 g/mol  x 0.9408  moles= 16.93  grams


3 0
3 years ago
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