Answer:
V = 0.24 L.
Explanation:
Hello there!
In this case, it is possible to approach this problem by using the ideal gas equation defined by:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
Whereas our unknown is V, volume, and we solve for it as follows:
![V=\frac{nRT}{P}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D)
Then since the STP conditions are ate 273.15 K and 1 atm, the volume of 0.011 moles of CO2 gas turns out to be:
![V=\frac{0.011mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=0.24L](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B0.011mol%2A0.08206%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A273.15K%7D%7B1atm%7D%5C%5C%5C%5CV%3D0.24L)
Regards!
<u>Answer: </u>
The pressure of a sample of argon gas was increased from 2.12atm to 6.96atm at constant temperature. If the final volume of the argon sample was 16.9L
. The initial volume of the argon sample was 55.5L
<u>Explanation:</u>
Considering the initial volume, initial pressure to be V1 and P1 respectively
Assuming the final volume and final temperature of the gas to be V2 and P2 respectively
Given,
Pressure of the gas was increased from 2.12 atm to 6.96 atm
Therefore, P1 = 2.12 atm
P2 = 6.96 atm
Final volume, V2 = 16.9 L
Applying Boyle’s law,
![P1\times V1 = P2\times V2](https://tex.z-dn.net/?f=P1%5Ctimes%20V1%20%3D%20P2%5Ctimes%20V2)
Substituting the value
![2.12 \times (V1) = 6.96\times 16.9](https://tex.z-dn.net/?f=2.12%20%5Ctimes%20%28V1%29%20%3D%206.96%5Ctimes%2016.9)
![V1 = \frac{(6.96 \times 16.9)}{2.12}](https://tex.z-dn.net/?f=V1%20%3D%20%5Cfrac%7B%286.96%20%5Ctimes%2016.9%29%7D%7B2.12%7D)
V1 = 55.5 L
Therefore, the initial volume was 55.5L
Answer
i like it when kids rub my legs
Explanation:
bernie sanders
<h3><u>Answer;</u></h3>
A. density.
High- and low-pressure areas leading to air movement happen when there are differences in <u>density</u>.
<h3><u>Explanation;</u></h3>
- <em><u>Atmospheric pressure or air pressure is the force exerted by air above a surface as gravity pulls it to Earth.</u></em>
- Air pressure decreases with an increase in altitude. Atmospheric pressure is measured using a barometer and is an important indicator of weather.
- <em><u>When a low-pressure system moves into an area, it causes cloudiness, wind, and precipitation. High pressure systems on the other hand causes a fair, calm weather.</u></em>