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Anit [1.1K]
4 years ago
5

Does the mass of the sealed jar and its contents change upon the vaporization of the liquid?

Chemistry
1 answer:
Norma-Jean [14]4 years ago
4 0

The answer is No. That is the mass of the sealed jar and its contents does not change upon the vaporization of the liquid, as according to the “law of conservation of mass” , the mass remain conserved when no matter is escape, the mass will remain constant and here also as the jar is sealed, no matter is escaped thus no mass change will be there.

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A glass blower can bend and shape a piece of glass that has been heated. Is glass a crystalline or an amorphous solid?
zavuch27 [327]

it would be a crystalline solid, because it could be extended in multiple directions.

5 0
3 years ago
If the moon resembled an asteroid, which theory would be most supported?
Law Incorporation [45]
<span>This is called "The capture theory" which states that the Earth's Moon was captured by the gravitational pull of our planet, meaning that it formed elsewhere, and then was pulle dinto place by the Earth, already formed. This would rely upon the ideas that a asteroid pased close enough to the Earth's orbit to be "captured".</span>
6 0
3 years ago
A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular
Gelneren [198K]

Answer:

Explanation:

Partial pressure of oil = mole fraction of oil x total pressure

mole fraction of oil = mole of oil / mole of water + mole of oil

= mole of oil = mass of oil / molecular weight of oil

= 20 / 100 = .2

mole of water = 80 / 18

= 4.444

mole fraction of oil =  .2 / .2 + 4.444

= .2 / 4.644

Partial pressure of oil = mole fraction of oil x total pressure

= (.2 / 4.644 ) x 760 mm

= 32.73 mm Hg .

3 0
4 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
4 years ago
What would be the new volume of a gas if 455 of that gas is cooled from 30.0°C to -80.0°C?
Lapatulllka [165]

Answer:

44555

Explanation:

is the answer

5 0
3 years ago
Read 2 more answers
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