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Gre4nikov [31]
3 years ago
8

Purple Group Header Warm-Up Why can I smell a chocolate bar even though it is a solid? Read the description of the office myster

y and write your prediction below. On a warm afternoon, a Universal Space Agency office worker returned from lunch break and noticed the office smelled like chocolate. She didn't see anything in the air, but she noticed her coworker was opening the wrapper of a solid bar of chocolate. Why can the office worker smell the chocolate bar even though it is in solid form?
Chemistry
1 answer:
likoan [24]3 years ago
4 0

Answer:

The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas

Explanation:

There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.

These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.

For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.

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Ksivusya [100]

The number of moles of argon that must be released in order to drop.

Solution:

Initial Temperature = 25°c = 298 K

Final Temperature =125 °c = 398 K

Initial Moles (n1) = 0.40 mole

Now,  Using the ideal gas law,

n1T1 = n2T2

0.400×298 = n2 × 398

n2 = 0.299 mol

Moles of Argon released

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= 0.100 mol.

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A container initially holds 1.24mol of hydrogen gas and has a volume of 27.8L. Hydrogen gas was added to the container, and the
Fantom [35]

Answer:

After increasing the volume, we have 1.81 moles of hydrogen gas in the container

Explanation:

Step 1: Data given

Number of moles hydrogen gas (H2) = 1.24 moles

Volume of hydrogen gas (H2° = 27.8 L

The final volume is increas to 40.6 L

Step 2: Calculate the new number of moles

V1/n1 = V2/n2

⇒with V1 = the initial volume = 27.8 L

⇒with n1 = the initial number of moles H2 = 1.24 moles

⇒with V2 = the final volume = 40.6 L

⇒with n2 = the new number of moles = TO BE DETERMINED

27.8L / 1.24 moles = 40.6 L / n2

n2 = 40.6 / (27.8/1.24)

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