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2 years ago

What is the pressure of the ideal gas when the volume is decreased to 4.0 L and the temperature is increased to 265 K?

1 answer:

8 0

It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT. First, we solve the amount of the gas in moles using the said equation and the first conditions.

(2.0 atm) (5.0 x 10^3 cm^3) = n (82.0575 atm.cm^3/mol.K)(215 K)
n=0.5668 mol

Using the second conditions given, we obtain the new pressure.
P (4.0 x 10^3) = 0.5668 x <span>82.0575 x 265
P= 3.08 atm</span>

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When two moles of ethane react completely with oxygen, how many moles of carbon dioxide will be produced?A. 2.B. 4.C. 8.D. Unkno

babymother [125]

Answer:

Option B.

Explanation:

As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.

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2 years ago

What are “condensation nuclei”?

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Answer:

Cloud condensation nuclei or CCNs are small particles typically 0.2 µm, or 1/100 the size of a cloud droplet on which water vapor condenses. Water requires a non-gaseous surface to make the transition from a vapour to a liquid; this process is called condensation.

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2 years ago

How is condensation explained by the Kinetic Molecular Theory? Cooling reduces particle motion, resulting in coalescence by attr

lara31 [8.8K]

Temperature means, in this context, movement.

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7 0

3 years ago

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Could someone explain how they got this answer, explain step by step plz

gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

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3 years ago

Which sub-atomic particle is NOT found at the center of the atom?

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3 years ago

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