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Marina86 [1]
3 years ago
11

What are two characteristics of an ideal gas that are not true of a real gas

Chemistry
1 answer:
MrRissso [65]3 years ago
5 0

There are many differences between ideal gas and real gas; some of the main differences are as following:

  • An ideal gas follows the formula PV=nRT but a real gas does not always follow this formula.
  • There is no attraction between the molecules of an ideal gas. A real gas has significant particle attractions.
  • The particles of an ideal gas lose no energy to its container. A real gas conducts and radiates heat, thereby losing energy.
  • An ideal gas is infinitely compressible, a real gas will condense to a liquid at some pressure.
  • Real gas particles have a volume and ideal gas particles do not.
  • Real gas particles collide in-elastically (loses energy with collisions) and ideal gas particles collide elastically.


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Secretary of State John Hay sent his Open Door
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Secretary of State John Hay sent his Open Door Notes (1899-1900) to world powers to protect United States trading interests in China. The correct option among all the options that are given in the question is option "3". These notes were mainly in regards to providing equal opportunity for trades in China and also respecting China's sovereignty, territorial integrity and administration. The principles of operating in China was the same for the British as well as for the United States, but John Hay was the first person to give it a written form. After this the official policy of the United States was based on the written document during the first half of the 20th century.
3 0
3 years ago
Consider the following system at equilibrium:A(aq)+B(aq) <---> 2C(aq)Classify each of the following actions by whether it
velikii [3]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

  • On addition of reactant at equilibrium shifts the equilibrium in forward direction.
  • On addition of product at equilibrium shifts the equilibrium in backward direction.
  • On removal of reactant at equilibrium shifts the equilibrium in backward direction.
  • On removal of product at equilibrium shifts the equilibrium in forward direction.

A(aq)+B(aq)\rightleftharpoons 2C(aq)

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling A and halving B, equilibrium constant of the reaction = K'

K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling B and C, equilibrium constant of the reaction = K'

K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

5 0
3 years ago
Electrons can jump between energy levels but can never be found in what energy level
yawa3891 [41]

Answer:

Electrons can jump from energy level to energy level (for example energy level 1 to 2) but they can NEVER be found in between energy levels.

6 0
3 years ago
You carefully weigh out 13.00 g of CaCO3 powder and add it to 52.65 g of HCl solution. You notice bubbles as a reaction takes pl
Sedbober [7]
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

n(CaCO₃)=m(CaCO₃)/M(CaCO₃)
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Δm=13.00+52.65-60.32=5.33 g

m(CO₂)=5.33 g
n(CO₂)=5.33/44.01=0,1211 mol

w=0.1211/0.1299=0,9323 (93.23%)

7 0
3 years ago
Read 2 more answers
A chemist has a 3.25-mole sample of potassium (K) to use in an experiment. What is the mass of this sample?
s344n2d4d5 [400]
Hi!
• For this you want to convert moles to grams
• To do this you would simply multiple your moles times the atomic mass of potassium
3.25 x 39.098= 127.0685
• So your answer for this should be about 127.07 depending on how you need to round!
5 0
3 years ago
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