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Alja [10]
2 years ago
6

Part i. you considered the equilibrium: fe3+(aq) + scn-(aq) fe(scn)2+(aq), and compared the color of the solutions in four test

tubes. in the third test tube, you added 6 m naoh dropwise. how do you explain the formation of a precipitate and the shift in equilibrium in terms of lechatelier's principle?
Chemistry
1 answer:
Alexus [3.1K]2 years ago
3 0

Answer:

The equilibrium will be shifted to lift with the formation of a brown gelatinous precipitate of Fe(OH)₃.

Explanation:

  • Le Chatelier's principle states that <em>"when any system at equilibrium for is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract the effect of the applied change and a new equilibrium is established that is different from the old equilibrium"</em>.
  • The addition of NaOH will result in the formation of Fe(OH)₃ precipitate which has a brown gelatinous precipitate.
  • The formation of this precipitate cause removal and decrease of Fe³⁺ ions.
  • According to Le Chatelier's principle, the system will be shifted to lift to increase Fe³⁺ concentration and reduce the stress of Fe³⁺ removal and readjust the equilibrium again. So, the [Fe(SCN)²⁺] decreases.
  • Increasing [Fe³⁺] will produce a yellow color solution that contains a brown gelatinous precipitate of Fe(OH)₃.
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vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

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3 years ago
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
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CO 2 + H2O ==> H2CO3

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CO 2 + NaOH ==> NaHCO3 (Sodium BiCarbonate )

NaHCO3 + NaOH ==> Na2CO3 (Sodium Carbonate) + H2O

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