Ethanol (c2h6o) is a common intoxicant and fuel produced from the fermentation of various grains. how many moles of ethanol are
1 answer:
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The new volume of the hydrogen gas in the balloon is 1.5 L.
<u>Explanation:</u>
At STP, Temperature, T1 is 0° C = 0 + 273 K = 273 K
Pressure, P1 = 1 atm
Here given volume, V1 = 3.60 L
Balloon at sea,
Pressure, P2 = 2.5 atm
Temperature, T2 = 10° C = 10 + 273 = 283 K
Volume, V2 = ?
Here, we have to use the equation,
We have to rearrange the equation to get V2 as,
Now plugin the values as,
V2 =
= 1.5 L
So the new volume of the hydrogen gas is 1.5 L.
Answer:
0.228 L is the volume that the gas occupies after it is driedand stored at STP.
Explanation:
Pressure of the wet gas =
1 atm = 760 Torr
Pressure of water vapor = p = 0.0372 atm
Pressure of gas =
Volume of wet gas =
1 mL = 0.001 L
Temperature of the wet gas =
Pressure of dry gas at STP =
Temperature of the dry gas at STP ,
Volume of dry gas at STP =
Using combined gas law:
0.228 L is the volume that the gas occupies after it is driedand stored at STP.
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x
=
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
Learn more about molarity here:
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