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s344n2d4d5 [400]
3 years ago
13

A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated

. b. use of an inert carrier gas. c. use of a mobile and a stationary phase. d. use of molecules that are soluble in water.
Chemistry
1 answer:
podryga [215]3 years ago
4 0

Answer : The correct option is, (c) use of a mobile and a stationary phase.

Explanation :

Chromatography : It is a separation process or technique of a mixture in which a mixture is distributed between the two phases at different rates, one of which is stationary phase and another is mobile phase.

Mobile phase : The mixture is dissolved in a solution is known as mobile phase.

Stationary phase : It is an adsorbent medium and It is a solid, liquid or gel that remains immovable when a liquid or a gas moves over the surface of adsorbent. It remains stationary.

Hence, a characteristic feature of any form of chromatography is the use of a mobile and a stationary phase.

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Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
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Answer:

This is the balanced equation:

Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq)  +  PbI₂ (s) ↓    

Explanation:

This are the reactants:

PbNO₃

NaI

Iodide can react to Pb²⁺ to make a solid compound.

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Which change does an increase in heat energy cause?
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C more collision between molecules

Explanation:

increase in heat causes increase in kinetic energy of the particles

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PLEASE HELP ME ASAP!
uranmaximum [27]

your answer is b hope this helps


7 0
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What happen when you add heat to solids and liquids?​
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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
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