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s344n2d4d5 [400]
2 years ago
11

Rank the following solutes in order of increasing entropy when 0.0100 moles of each dissolve in 1.00 liter of water.

Chemistry
1 answer:
vichka [17]2 years ago
4 0

Answer:

<em>The rank of the solutes in order of increasing entropy is</em>:

  •   <em>CH₃OH < NaBr < CaCl₂ < Cr(NO₃)₃</em>

Explanation:

Assuming temperature and other conditions are the same, you can qualititatively precict the relative increase of entropy, when a given amount of solute dissolves in a fixed aomount of water, by looking at the number of particles (molecules or ions) that are formed.

The particles of solute that, prior to get dissolved, are pure and orderly separated, will be very scattered after the dissolution, increasing, consequently, the degree of randomness or dissorder of the system.

Also, the greater the number of particles formed from each unit formula, the greater the degree of disorder and the greater the entropy.

Hence, being the entropy a measure of the radnomness of the system, you can predict that the greater the number of particles (molecules or ions) dissolved, the greater the entropy.

Now look at the dissociation of each given solute:

  • CaCl₂ → Ca²⁺ + 2Cl⁻ : 3 ions are formed.

  • CH₃OH: is a molecular compound, so 1 molecule is dispersed.

  • Cr(NO₃)₃ → Cr³⁺ + 3 NO₃⁻: 4 ions are formed.

  • NaBr → Na⁺ + Br⁻ : 2 ions are formed.

Then, the rank in order of increasing entropy, is:

       1       <     2   <     3     <      4

  CH₃OH < NaBr < CaCl₂ < Cr(NO₃)₃

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What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places
lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

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Answer:

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