The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.
<h3>What is
Boyle's Law?</h3>
According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.
For the given question we use the below equation is:
P₁V₁ = P₂V₂, where
P₁ = initial pressure of gas = 1.5 atm
V₁ = initial volume of gas = 2 L
P₂ = final pressure of gas = ?
V₂ = final volume of gas = 1 L
On putting all these values on the above equation, we get
P₂ = (1.5atm)(2L) / (1L) = 3 atm
Hence required pressure of the gas is 3 atm.
To know more about Boyle's Law, visit the below link:
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Explanation:
- As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.
As in
, the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.
Hence, these bonds are so strong that high heat needs to given to break them.
- A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.
Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.
- In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.
As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.
Answer:

Explanation:
The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.
So, we need to calculate the mass of the acetic acid as follows:

Where:
n: is the number of moles = C*V
M: is the molecular mass = 60.052 g/mol
C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L
V: is the volume = 50 ml = 0.050 L

Now, the amount of solute adsorbed per gram of the adsorbent is:

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.
I hope it helps you!