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ioda
3 years ago
9

Min increases the temperature of a gas in an expandable container. If she keeps the pressure constant, what will happen to the v

olume of the gas?
Chemistry
2 answers:
Galina-37 [17]3 years ago
8 0

Answer:

The volume will remain the same.

Explanation:

The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's law)

avanturin [10]3 years ago
8 0

Answer:

It would remain the same ig..

Explanation:

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5 0
2 years ago
At 920 K, Kp = 0.40 for the following reaction. 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium p
Alex777 [14]

<u>Answer:</u> The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

<u>Explanation:</u>

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

                        2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

<u>Initial:</u>                 0.52      0.52

<u>At eqllm:</u>         0.52-2x    0.52-x        2x

The expression of K_p for above equation follows:

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}

We are given:

K_p=0.40\\\\p_{SO_3}=2x\\\\p_{SO_2}=0.52-2x\\\\p_{O_2}=0.52-x

Putting values in above equation, we get:

0.40=\frac{(2x)^2}{(0.52-2x)^2\times (0.52-x)}\\\\x=-1.13,-0.402,0.077

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide = (0.52-2x)=(0.52-(2\times 0.077))=0.366atm

Equilibrium partial pressure of oxygen = (0.52-x)=(0.52-0.077)=0.443atm

Equilibrium partial pressure of sulfur trioxide = 2x=(2\times 0.077)=0.154atm

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

4 0
3 years ago
How was the speed of light determined and what is it’s value?
zheka24 [161]

The speed of light in a vacuum stands at “exactly 299,792,458 metres per second“. The reason today we can put an exact figure on it is because the speed of light in a vacuum is a universal constant that has been measured with lasers; and when an experiment involves lasers, it's hard to argue with the results.

5 0
3 years ago
Why aren't descriptive investigations repeatable?
Ipatiy [6.2K]
Descriptive investigations are mainly questions to uncover information.  it  does not have an hypothesis.  It is mainly gathering information, it is not repeatable as you are just asking questions

3 0
3 years ago
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Can you help me please ​
9966 [12]

Answer:

I think its, balanced chemical equations

Explanation:

8 0
3 years ago
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