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aalyn [17]
2 years ago
10

1) A 3dm vessel can withstand a maximum internal pressure of 35 atmosphere, what is the highest temperature it can be heated to,

if 4 moles of nitrogen gas is pumped into the vessel?​
Chemistry
1 answer:
matrenka [14]2 years ago
4 0

Answer:

hbffbb bbfgh vgdggjudty

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A sample of xenon gas at a pressure of 1.14 atm and a temperature of 20.4 °C, occupies a volume of 16.9 liters. If the gas is al
kondor19780726 [428]

Answer:

P2= 0.696atm

Explanation:

Applying Boyle's law

P1V1= P2V2

1.14×16.9= P2× 27.7

Simplify

P2= (1.14×16.9)/27.7

P2= 0.696atm

8 0
3 years ago
What is the pH of a solution with a hydrogen ion concentration of 4.5 x<br> 10-6 M ?
Fittoniya [83]

Answer:

<h2>5.35 </h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

We have

pH =  -  log(4.5 \times  {10}^{ - 6} )  \\  = 5.346787...

We have the final answer as

<h3>5.35 </h3>

Hope this helps you

6 0
3 years ago
Identify the substance that is oxidized. Copy the oxidation half reaction for this substance.
slava [35]

please check the attached image, i have given this answer b4

4 0
3 years ago
An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the am
gogolik [260]

Answer:

Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

Explanation:

4 0
3 years ago
Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days
nadya68 [22]

Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

Suppose the original amount is 100.00 grams

Step 2: Calculate the percentage left

X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

X = 54.29

There is 54.29 % sample left after 12.6 days

8 0
3 years ago
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