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AURORKA [14]
3 years ago
14

PLEASE PLEASE HELP ME WITH THIS ASAP

Chemistry
1 answer:
Pavel [41]3 years ago
5 0

its letter b i guess

change my ans if u think its wrong

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The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction
shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

       NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)

(a)  When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of H_{2}S will increase.

  • (b)  When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of H_{2}S will decrease with decrease in volume.

  • When we increase the mount of NH_{4}HS then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of H_{2}S will increase then.

3 0
3 years ago
What is the difference between ionization energy and ionization potential​
rjkz [21]

Answer:

They are similarly

Explanation:

3 0
3 years ago
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo
siniylev [52]

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

3 0
3 years ago
PLEASE HELP ASAP!!!!! WILL GIVE BRAINLIEST!!!!
Akimi4 [234]

Explanation:

The magnet uses electromagnetic induction meaning it can be readily magnetized and demagnetized (using electricity) when required. When electricity is switched on, it becomes magnetized and lifts an object and when electricity is switched off, it loses magnetism and releases the object.

The magnetic is able to lift heavy objects because it has powerful conductor material (ferromagnetic iron) and the number of electromagnetic coils is many to induce a powerful magnetic force. The electric current, inducing the magnetism, is also powerful.

Learn More:

For more on electromagnetic induction check out;

brainly.com/question/3414535

brainly.com/question/13369951

#LearnWithBrainly

7 0
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).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
4 years ago
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