Answer:
yes
Explanation:
because I think it is correct fo u
The compound that would have the highest osmotic pressure when dissolved in water is
.
So, option D is correct one.
The dissociation of one molecule of
gives the maximum number of ions when dissolved in water ( 4 ions ) . Osmotic pressure is a colligative property and depends upon number of solute particles present in the solution . The solution having maximum number of solute particles will have maximum number of the osmotic pressure .
All other given molecules gives less number of number of ions when dissolved in water as compare to of
.
To learn more about osmotic pressure
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Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

Dividing the previous equation by x:

We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

Calculate the number of moles of CO2 and water considering the same:


The total number of moles at the reactor output would be:

So, the oxygen mole fraction would be:
%
= k
<u>Explanation:</u>
The relation between volume, V of gas and Temperature, T of a gas is related by Charles Law.
This law states that the volume of a given amount of gas held at a constant pressure is directly proportional to the Kelvin temperature
Thus,
= k
where k is a constant
Therefore,
=
=
...
This shows, as the volume of a gas goes up, the temperature also goes up and vice-versa.