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
sdas [7]
3 years ago
6

When the partition is removed, the gas expands and fills both compartments. How many moles n of gas were initially contained in

the lower compartment if the entropy change of the gas in this free-expansion process is 17.28 J/K
Chemistry
1 answer:
Tatiana [17]3 years ago
4 0

Answer:

n = 1.89 mol.

Explanation:

Think of this as adiabatic free expansion. This means constant temperature -- we know this because the internal energy does not change.

Since the temperature is constant, the entropy change can be given by:

Entropy = nRln(V2/V1)

Entropy = 17.28 J/K. R = 8.314 (gas constant). V2 = 3V. V1 = V.

Thus, we get:

17.28 = n(8.314)ln3

Solve for n. We get n = 1.89 mol.

You might be interested in
What type of bond occurs between tow nonmetals.
Aleksandr-060686 [28]

Answer:

covalent bonds............

6 0
2 years ago
Read 2 more answers
4Fe + 30₂ ⇒ Fe₂0₃
DaniilM [7]

Answer:

A

The nuber of each one should be same

5 0
2 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?
Arlecino [84]
2KClO₃ → 2KCl + 3O₂

mole ratio of KClO₃ to O₂ is 2 : 3

∴ if moles of O₂ = 5 mol

then moles of KClO₃ = \frac{5 mol   *   2}{3}

                            = 3.33 mol


Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃

                                      = 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol

                                      = 407.93 g
6 0
3 years ago
How many grams of O are in 905 grams of Na2O? ​
galben [10]

Answer:

301.7g

Explanation:

So there's 1 out of 3 are Oxygen.

905 / 3 = 301.7

8 0
3 years ago
Other questions:
  • Which of the following states that an orbital can contain two electrons only if all other orbitals at that sublevel contain at l
    7·1 answer
  • The information below describes a redox reaction.
    8·1 answer
  • Name and explain the two types of atom bonding
    6·1 answer
  • Explain why organisms are more likely to be well preserved in mud than in sand
    11·1 answer
  • The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a
    12·1 answer
  • If ℓ= 1, what can you deduce about n?
    8·1 answer
  • Waves that require a medium through which to travel are electromagnetic waves. True or False?
    8·2 answers
  • What species is reduced in the reaction below?
    15·1 answer
  • I run the reaction below using 55.0 g sodium chloride and plenty of fluorine, and it produces 8.8 grams of chlorine gas. What is
    14·1 answer
  • co3 forms a more stable complex with the chelating ligand en (h2nch2ch2nh2) than it does with the monodentate ligand methylamine
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!