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denis23 [38]
3 years ago
8

Each of the given statements describes a type of column chromatography. Match the statements to the type of chromatography they

describe. If a statement can describe all of the types, place that statement in the All category. (Note that size‑exclusion chromatography may also be called gel filtration or molecular‑exclusion chromatography.)
Categories: Size-exclusion Chromatography, Affinity Chromatography, Ion-exchange Chromatogrpahy, and All

Statements:
(a) Separate molecules by size.
(b) Separate molecules by charge.
(c) The stationary phase has a covalently bound group to which a protein in the mobile phase can bind.
(d) Uses a mobile phase and a stationary phase to separate proteins.
(e) The stationary phase contains cross-linked polymers with different pore sizes.
(f) Can separate molecules based on protein-lignand binding.
(g) The stationary phase may contain negatively or positively charged groups.
Chemistry
1 answer:
vlada-n [284]3 years ago
6 0

Answer:

The proper matching is given below.

Explanation:

a Separate molecules by size    size exclusion chromatography

b Separate molecules by charge Ion exchange chromatography

c The stationary phase has a covalently bound group to which a protein in the mobile phase can bind.                  Affinity chromatography

d uses mobile phase and stationary phase to separate protein Size exclusion chromatography

e The stationary phase contain  cross linked polymers with different pore size

                                                                  Size exclusion chromatography

f can separate molecules based on protein ligand binding   Affinity chromatography

g The stationary phase may contain negatively or positively charged groups

  ion exchange chromatography            

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Explanation:

The reaction of calcium bromide with lithium oxide will be:

CaBr_2+Li_2O\rightarrow 2LiBr+CaO

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of calcium bromide}=\frac{2.5g}{199.9g/mol}=0.0125moles

As lithium oxide is in excess, calcium bromide is the limiting reagent.

According to stoichiometry :

1 mole of CaBr_2 produce = 2 moles of LiBr

Thus 0.0125 moles of CaBr_2 will require=\frac{2}{1}\times 0.0125=0.025moles  of NH_3

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