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8090 [49]
3 years ago
9

The heat of vaporization of an unknown liquid is 1,340 Joules per gram. What is the minimum number of Joules needed to change 20

.0 grams of the liquid to vapor at the boiling point?
Chemistry
1 answer:
Fed [463]3 years ago
7 0

Answer:

The temperature of a boiling liquid remains constant until all of the liquid has been converted to a gas.For example, the amount of heat necessary to change one gram of water to steam at its boiling point at one atmosphere of pressure,  the heat of vaporization of water, is approximately 540 calories.

Explanation:

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Consider two equal-volume balloons under the same conditions of temperature and pressure. One contains helium, and the other con
Olegator [25]

Answer:

1. Number of gas particles (atoms or molecules)

2. Number of moles of gas

3. Average kinetic energy

Explanation:

Since the two gas has the same volume and are under the same conditions of temperature and pressure,

Then:

1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole

2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.

3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.

5 0
3 years ago
Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)
Gelneren [198K]

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

                              Fe₂O₃ +  3C →  2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

      ΔS = ∑S_{products} - ∑S_{reactants}

Therefore:

     ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

                          ΔS = +541.3Jmol⁻¹K⁻¹

5 0
3 years ago
Please i need help really quickly on this. thank you​
prohojiy [21]

Answer:

the error could have been the fact that the unit for volume wasn't changed from cm³ to dm³

hence the calculation error

the solution to this would be first dividing the volume by 1000 to get that same amount in dm³ which is the standard unit to be used for volume-density calculations

6 0
3 years ago
How to balance reaction equations
spayn [35]
They will become unequivalent

7 0
3 years ago
If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?
GarryVolchara [31]

Answer:

The amount in grams of hydrogen gas produced is 0.551 grams

Explanation:

The parameters given are;

Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms

Chemical equation for the reaction is given as follows;

2K + 2H₂O \rightarrow KOH + H₂

Avogadro's number, N_A, regarding the number of molecules or atom per mole is given s follows;

N_A = 6.02 × 10²³ atoms/mole

Therefore;

The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles

2 moles of potassium produces one mole of hydrogen gas, therefore;

1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas

The molar mass of hydrogen gas = 2.016 grams

Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.

The amount in grams of hydrogen gas produced = 0.551 grams.

8 0
3 years ago
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