1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mamaluj [8]
3 years ago
6

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

29 76.02 205.14 (J · K −1 · mol−1 ) at 298 K. 1. −117.1 kJ · mol−1 2. +246.2 kJ · mol−1 3. −64.5 kJ · mol−1 4. +117.1 kJ · mol−1 5. −246.2 kJ · mol−1
Chemistry
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

You might be interested in
Which statement is true for a cooling curve​
rjkz [21]

what are your choices

6 0
3 years ago
A tire at 21°c has a pressure of 0.82 atm. its temperature decreases to –3.5°c. if there is no volume change in the tire, what i
rjkz [21]

The pressure of the tyre after the temperature change is 0.89atm. Details about pressure can be found below.

<h3>How to calculate pressure?</h3>

The pressure of the tyre can be calculated using the following equation:

P1/T1 = P2/T2

Where;

  • P1 = initial pressure
  • P2 = final pressure
  • T1 = initial temperature
  • T2 = final temperature

According to this question, a tyre at 21°C (294K) has a pressure of 0.82 atm. Its temperature decreases to –3.5°C (269.5K).

0.82 × 294 = P2 × 269.5

241.08 = 269.5P2

P2 = 241.08 ÷ 269.5

P2 = 0.895atm

Therefore, the pressure of the tyre after the temperature change is 0.89atm.

Learn more about pressure at: brainly.com/question/23358029

#SPJ4

6 0
2 years ago
Which event is an example of melting?
avanturin [10]

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

8 0
3 years ago
An environmental scientist studies _______.
Assoli18 [71]

An environmental scientist studies the environment - you can see that in the beginning of the sentence :)
7 0
3 years ago
Read 2 more answers
Two students made the Lewis dot diagrams of H2O. The diagrams are as shown.
netineya [11]

Answer:

Only student A

Explanation:

don’t worry I got a 100

7 0
3 years ago
Read 2 more answers
Other questions:
  • Type one to two paragraphs describing the changes in potential and kinetic energy of the cart.
    6·2 answers
  • What is a substance that decreases the rate of a chemical reaction called?
    14·2 answers
  • In the reaction H2SO4 + 2 NaOH -&gt; Na2SO4 + 2H2O, an equivalence point occurs when 23.1 mL of 0.2055 M NaOH is added to a 25.0
    10·1 answer
  • Find the volume of 20.0 g of benzene
    7·1 answer
  • what type of wave uses thioglycolic acid or its derivatives with ammonia and procceses the hair without heat?
    14·1 answer
  • 1. In a Stock system name such as iron(III) sulfate, the Roman numeral tells us (a) how many atoms of Fe are in one formula unit
    9·1 answer
  • Round off 35.65 to three significant figures.<br><br><br> help
    12·1 answer
  • What's the agriculture​
    7·2 answers
  • Which green house gas is produced by commercial refrigeration and air conditioning systems
    8·1 answer
  • Potassium carbonate (K2CO3) has one potassium atom bonded to carbon, and that carbon is also bonded to three oxygens. Their elec
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!