D. dormancy is the correct answer
Answer:
-Unknown
Explanation:
Questions is not seen properly bro
All the information is answered/given in the paragraph except that the study doesn’t show cells in their natural habitat!
Question:
At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?
Answer:
The new pressure is: 1.896 atm
Explanation:
At standard temperature and pressure, we have:
![P_1 = 1atm](https://tex.z-dn.net/?f=P_1%20%3D%201atm)
![T_1 = 273.15k](https://tex.z-dn.net/?f=T_1%20%3D%20273.15k)
![V_1 = 3.5L](https://tex.z-dn.net/?f=V_1%20%3D%203.5L)
Outside, we have:
![T_2 = 296k](https://tex.z-dn.net/?f=T_2%20%3D%20296k)
![V_2 = 2.0L](https://tex.z-dn.net/?f=V_2%20%3D%202.0L)
Required
Determine the new pressure
Using combined gas law, we have:
![\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%20%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
This gives:
![\frac{1 * 3.5}{273.15} =\frac{P_2*2.0}{296}](https://tex.z-dn.net/?f=%5Cfrac%7B1%20%2A%203.5%7D%7B273.15%7D%20%3D%5Cfrac%7BP_2%2A2.0%7D%7B296%7D)
Solve for ![P_2](https://tex.z-dn.net/?f=P_2)
![P_2 = \frac{296 * 1 * 3.5}{273.15*2.0}](https://tex.z-dn.net/?f=P_2%20%3D%20%5Cfrac%7B296%20%2A%201%20%2A%203.5%7D%7B273.15%2A2.0%7D)
![P_2 = \frac{1036}{546.30}](https://tex.z-dn.net/?f=P_2%20%3D%20%5Cfrac%7B1036%7D%7B546.30%7D)
![P_2 \approx 1.896 atm](https://tex.z-dn.net/?f=P_2%20%5Capprox%201.896%20atm)