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Maslowich
3 years ago
15

On increasing which of the following factors, decreases the extent of physisorption? Surface area of the adsorbent Critical temp

erature of the adsorbate Concentration of adsorbate Temperature at which adsorption is carried out
Chemistry
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer: On increasing temperature at which adsorption is carried out decreases the extent of physisorption.

Explanation:

An adsorption where molecules of the adsorbate are placed or held on the surface of adsorbent by Vander waals forces is called physisorption.

There is basically physical bonding between the molecules of gas to the surface of a solid or liquid.

Physisorption is reversible in nature and occurs at low temperatures.

It is not specific in nature which means that all gases are adsorbed on the surface of every solid substance to some extent.

Thus, we can conclude that on increasing temperature at which adsorption is carried out decreases the extent of physisorption.

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A measurement 172.54 m rounded to three significant figures is
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2 years ago
Explain why, when the imidazole ring of histidine is protonated, the double-bonded nitrogen is the nitrogen that accepts the pro
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4 0
3 years ago
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
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