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ad-work [718]
3 years ago
9

Plz help it's due in 5 min​

Chemistry
2 answers:
jonny [76]3 years ago
8 0

Answer:

51.84

Explanation:

GarryVolchara [31]3 years ago
7 0

Answer:

what is the question then we can help you out

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Sound waves are transverse mechanical waves. <br><br> True<br> False
allochka39001 [22]

False

Explanation:

Sound waves are longitudinal mechanical waves.

Mechanical waves are waves that requires a medium to propagate them.

  • A transverse wave is a wave that is propagated perpendicularly to its source.
  • An example is electromagnetic radiation.
  • A longitudinal wave is a wave that is directed parallel to their source.
  • Sound wave is a longitudinal wave.
  • It has series of rarefaction and compression along its travel path.
  • Rarefaction are areas of sparse particles.
  • Compression are areas with dense particles.

learn more:

Sound wave brainly.com/question/3619541

#learnwithBrainly

7 0
3 years ago
How many moles of aluminum oxide al2o3 are in a sample with a mass of 204.0
erik [133]

Answer:

2 moles

Explanation:

Let us first start by calculating the molecular mass of Al₂O₃.

The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.

Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

Therefore, molecular mass of Al₂O₃ is:

= (2*26.981539) + (3*15.999) u

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

3 0
3 years ago
Read 2 more answers
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
5 0
3 years ago
Please hurry this is due tomorrow morning
Llana [10]
Answer: G Atom 1 and Atom 4.. hope this helps and good luck!
3 0
3 years ago
When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
3 years ago
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