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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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Read 2 more answers
The pH at the equivalence point will be a. less than 7.00 b. equal to 7.00 Allohalineup of a sto quivalence point of a titration
True [87]

Answer: c. greater than 7.00

Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.

8 0
2 years ago
Determine how many grams of N2 are produced from the reaction of 8.37 g of H2O2 and 5.29 g of N2H4.
Dmitriy789 [7]
N₂H₄  +  2H₂O₂    →   N₂  +  4H₂O

mol = mass ÷ molar mass

If mass of hydrazine (N₂H₄)  = 5.29 g 
then mol of hydrazine           = 5.29 g ÷ ((14 ×2) + (1 × 4))
                                              = 0.165 mol

mole ratio of hydrazine to Nitogen is     1   :  1
  ∴ if moles of hydrazine = 0.165 mol
     then moles of nitrogen = 0.165 mol

Mass = mol × molar mass

Since mol of nitrogen (N₂)  = 0.165
then mass of hydrazine      = 0.165 × (14 × 2)
                                           = 4.62 g
5 0
2 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
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