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iVinArrow [24]
4 years ago
6

Experiment predicted observation A student has two unopened cans containing carbonated water. Can A has been stored in the garag

e () and can B has been stored in the fridge (). The student opens one can at the time, both cans make a fizz.
A) The fizz will be the same for both cans
B) There is not enough information to predict which can will make the louder fizz
C) Can A will make a louder and stronger fizz than can B.
D) Can B will make a louder and stronger fizz than can A.
Chemistry
1 answer:
nlexa [21]4 years ago
5 0

Answer:

Can A will make a louder and stronger fizz than can B.

Explanation:

Temperature has a direct effect on gas solubility. We know that carbonated water contains carbon dioxide dissolved in water. The extent of dissolution or solubility of this gas is dependent on the temperature of the system.

As the temperature of the system rises, the solubility of gas in solution decreases. It follows that can A, having been stored in a garage is definitely at a higher temperature than can B stored in the refrigerator.

Since solubility of gases decreases with increasing temperature, the carbon dioxide in can A will be less soluble than in can B. This will cause can A to make a louder and stronger fizz when opened than can B.

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Answer:

V₂ = 1.41 L

Explanation:

Given data:

Initial temperature = 35°C (35 +273.15 K = 308.15 K)

Initial volume = 1.5 L

Final temperature = 17°C (17+273.15 K = 290.15 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

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Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1.5 L × 290.15 K / 308.15 k

V₂ = 435.23 L.K / 308.15 k

V₂ = 1.41 L

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