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
SashulF [63]
3 years ago
5

0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reache

d the pressure of CO2 is 0.220 atm. 0.300 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3
Chemistry
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

12.531 grams will be the final mass of calcium carbonate.

Explanation:

CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)

Partial pressure of carbon dioxide gas at equilibrium =P=0.220 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.220 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.06960 mol

According to reaction , 1 mole of carbon dioxide gas is formed from 1 mole of calcium carbonate,0.06960 mole of carbon dioxide gas will be obtained from :

\frac{1}{1}\times 0.06960 mol=0.06960 mol calcium carbonate

Moles of calcium carbonate at equilibrium = 0.100 mol - 0.06960 mol = 0.03040 mol

After addition of 0.300 atm of carbon dioxide gas, more amount of calcium carbonate will be be formed.

Here, at the same temperature, the equilibrium pressure of the carbon dioxide gas is 0.220 atm so, entire 0.300 atm of carbon dioxide will get convert to calcium carbonate.

So amount moles of carbon dioxide gas added = moles of calcium carbonate formed after re-establishment of an equilibrium :

Partial pressure of carbon dioxide gas added at equilibrium =P=0.300 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.300 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.09491 mol

Moles of calcium carbonate = 0.09491 mol

Total moles of calcium carbonate in the container :

=0.03040 mol + 0.09491 mol = 0.12531 mol

Mass of 0.12531 moles of calcium carbonate :

0.12531 mol × 100 g/mol = 12.531 g

12.531 grams will be the final mass of calcium carbonate.

You might be interested in
Which of the following statements is correct concerning the reaction 2 A + B → 2 C + 2 D?
Eddi Din [679]

Answer:

2 is the correct answer maybe

5 0
3 years ago
Describe the similarities and differences between herbivores,carnivore,and omnivores​
vredina [299]

Answer:

Herbivores only eat plants . Remember HERBS are PLANTS.

Carnivores only eat meat. Remember CARNE = MEAT in Spanish.

Omnivores eat MEAT & PLANTS.

Explanation:

Differences are what they eat, similarities omnivores eat both.

3 0
3 years ago
Why is salt water intrusion a big problem in florida?
Schach [20]

Its leading to contaminated drinking water.

7 0
3 years ago
Read 2 more answers
What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit
jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution  = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0
2 years ago
What is the molarity of a solution made by dissolving 1.25 mol of HCl in enough
Ronch [10]

Answer:

Explanation:

C

4 0
3 years ago
Other questions:
  • What are the characteristics of a continental drift?
    5·1 answer
  • How many mL of a 0.300 M AgNO3 solution will it take to make 500 mL of a 0.100 M AgNO3 solution?
    6·2 answers
  • If 3.11 mol of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 L, what is the temperature of the sample in degrees
    10·1 answer
  • With all the coefficient set to 1, how can you tell that the equation is not balanced?
    10·1 answer
  • Witch two chemists organized the elements based on properties suck as how the elements reacts or whether they are solid or liqui
    15·1 answer
  • An example of a physical change is
    9·2 answers
  • Which of the following statements is true?
    15·2 answers
  • Two mice have different parents. They both have the same protein for whisker thickness in their cells.
    10·1 answer
  • Ano ang scientific method​
    6·1 answer
  • A student has a test tube that contains 50 mL of water. She wants to add a substance to the water that will cause the test tube
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!