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Inessa05 [86]
3 years ago
7

Describe the interactions between molecules in a mixture (analyte), the stationary phase and various mobile phases used in colum

n chromatography and the effect of these interactions on the separation of the mixture.
Chemistry
1 answer:
ratelena [41]3 years ago
4 0
<h2>Hydrophobic interactions. </h2>

Explanation:

  • Column chromatography is a technique used for separation in which a mixture having a mobile phase is caused to move in contact with the stationary phase.
  • The Stationary phase used is a solid and the mobile phase used is liquid.
  • There are interactions between the molecules of the analyte, stationary phase, and the various mobile phases like hydrogen bonding and dipole-dipole interactions.
  • The effect of these interactions is that the compound is retained for a longer time and as there are different interactions between stationary and mobile phases they are carried along with the mobile phases with varying degrees and the separation of the mixture is done.
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Organisms can either be made of one cell or made of more than one cell
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Answer:

The statement is true.

Explanation:

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

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The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

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3 years ago
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