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Alexeev081 [22]
3 years ago
7

during which of the following processes does the entropy of the system decrease? salt crystals dissolve in water air escapes fro

m a hole in a balloon ice melts at room temperature liquid water boils methane gas and oxygen gas combust to form carbon dioxide gas and liquid water
Chemistry
1 answer:
Nat2105 [25]3 years ago
4 0
<h2>Entropy of the system decreases when methane gas and oxygen gas  combust to form carbon dioxide and water.</h2>

Explanation:

  • Salt crystals when dissolved in water the bond between the molecules of crystals are broken down increasing the movement between the atoms hence entropy is increased.
  • When air escapes from a hole in the balloon,the entropy increases as the air present inside the balloon occupies small volume as compared to outside.
  • Ice when melts at room temperature its entropy increases because the molecule in the ice are closely packed hence less movement between the molecules.
  • When water boils the kinetic energy increases or themovement between the atoms increases which increases the entropy.

Hence, the option (a), (b), (c), and (d) are not suitable to the statement                       that  "the entropy of the system decreases".

  • During combustion of methane gas there is release of carbon dioxide gas and water .Here, entropy decreases due to the formation of water which is due to overall decrease in the  kinetic energy of the system.
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The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions
musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

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