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fredd [130]
3 years ago
7

There are four distinct techniques used to separate mixtures described in this lesson. Name and describe each technique.

Chemistry
1 answer:
tino4ka555 [31]3 years ago
4 0
<h2>There are four distinct techniques used to separate mixtures that is : </h2>

Explanation:

Mixtures

They are formed when two or more substances are simply mixed in any ratio not chemically combined with each other.  

Characteristics  

  • They can be homogeneous or heterogeneous in nature that is the constituents can be seen to have visible boundaries or they may appear to mix thoroughly.
  • The Properties of mixtures are same as that of constituents.  
  • The constituents can be separated by physical methods.  
  • Their formations do not require or release energy as there is no bond formation or breakage involved.
  • The Properties of mixtures like melting point & boiling points are not fixed.  

Let us discuss few techniques :

1. Evaporation

For example :To obtain colored component (Dye) from Ink

Materials Required : Watch glass ink (blue/ black) beaker , stand, burner.  

Procedure :  

  • Take a beaker and fill it half with water.  
  • Take  ink (Blue/ black) in the watch glass and place it on the mouth of the beaker
  • Start heating the beaker and observe.  
  • Heating is continued as long as the evaporation is taking place
  • Heating is stopped when no any further change can be noticed on the watch glass  

Observation:  

Evaporation taking place from the watch glass can be seen  

Residue is left on the watch glass  

Conclusion drawn :

  • it can be concluded that ,”Ink is not a pure substance but it is a mixture of dye in water which can easily be separated by evaporation method”.  
  • Ink is not a single substance  

2, Centrifugal method

 For example : To separate cream from milk by using centrifuge method.  

Materials Required : Full cream milk centrifuging machine/milk churner, jug test-tubes.  

Procedure:  

  • Take un-boiled cold milk in two test-tubes and place these test-tubes in a centrifuging machine
  • Centrifuge it at high speed by using a hand centrifuging machine for two minute and observe.

Observation

Cream floating on the milk can be seen.  

Conclusion drawn

When milk is rotated at high speed then the suspended lighter particles (fats and protein molecules) bind with each other forming ‘cream’ and ‘skimmed milk’. The cream being lighter floats over the skimmed milk which can be removed easily.  

3. Sublimation

For example :To separate a mixture of common salt and ammonium chloride :  

Materials Required : China dish, tripod stand mixture of common salt and ammonium chloride glass funnel cotton and burner.  

Procedure:  

  • Take the mixture of sand and ammonium chloride in a china dish.  
  • Cover the china dish with an inverted glass funnel and place it on a tripod stand.  
  • Put a loose cotton plug in the opening of the funnel so as to prevent the escape of ammonium chloride vapors.  
  • Heat the china dish on a low flame and observe.  

Observation :  

  •  White fumes (vapors) of ammonium chloride can be seen coming out of the mixture.  
  • These white fumes start depositing as white solid on coming in contact with the cold, inner walls of the funnel.  
  • Sand salt is left behind in the china dish.

Conclusion drawn: Ammonium chloride, being a volatile substance , changes into white vapors easily which deposits on the cold inner wall of the funnel. This ammonium chloride obtained is called the  

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The decomposition of water produces 50g of Hydrogen gas and 75 g of oxygen gas. How much water
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84.24 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂O → 2  H₂ + O₂

Now we calculate the number of moles of products.

number of moles = mass / molar weight

number of moles of H₂ = 50 / 2 = 25 moles

number of moles of O₂ = 75 / 32 = 2.34 moles

We see from the chemical reaction that for every 2 moles of H₂ produced there are 1 mole of O₂ produces for every 25 moles of H₂ produced there are 12.5 moles of O₂ but we only have 2.35 moles of O₂ available. The O₂ will be the limiting quantity from which we devise the following reasoning:

if        2 moles of H₂O  produces 1 mole of O₂

then  X moles of H₂O  produces 2.34 mole of O₂

X = (2 × 2.34) / 1 = 4.68 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 4.68 × 18 = 84.24 g

Learn more about:

limiting reactant

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