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Softa [21]
2 years ago
12

A balloon is filled with 35.0 L of helium in the morning when the temperature is 303 degrees kelvin. By 3:00 p.m., the temperatu

re has risen to 333 degrees kelvin. What is the new volume of the balloon?
2882 L

38.5 L

31.8 L

65.0 L
Chemistry
2 answers:
Anna007 [38]2 years ago
7 0

Answer:

The new volume of the balloon is 38.5 L

Explanation:

Step 1: Data given

Volume at the start = V1 = 35.0 L

Temperature at the start = T1 = 303 Kelvin

Volume by 3pm = TO BE DETERMINED

Temperature by 3pm = 333 Kelvin

<u>Step 2: </u>Calculate the new volume

Charles' gas law says

V1/T1  = V2/T2

V 1  is the initial volume and T1 is the initial temperature

V2 is the final volume and T2 is the final temperature

35 L / 303 Kelvin = V2 / 333 Kelvin

V2 = 35L * 333 Kelvin / 303 Kelvin

V2 = 38.47L ≈ 38.5 L

The new volume of the balloon is 38.5 L

ladessa [460]2 years ago
4 0

Answer:

v=333/303 x 35.0= 38.5

Explanation:

volume goes up by the ratio of kelvin temp change

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In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.

Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.

For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.

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