This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.
Answer:
initial temperature=![-3.31^{\circ}C](https://tex.z-dn.net/?f=-3.31%5E%7B%5Ccirc%7DC)
Explanation:
Assuming that the given follows the ideal gas nature;
![P_1=0.723atm](https://tex.z-dn.net/?f=P_1%3D0.723atm)
![V_1=13.5L](https://tex.z-dn.net/?f=V_1%3D13.5L)
![T_1=?](https://tex.z-dn.net/?f=T_1%3D%3F)
![P_2=0.612atm](https://tex.z-dn.net/?f=P_2%3D0.612atm)
![V_2=17.8L](https://tex.z-dn.net/?f=V_2%3D17.8L)
![T_2=28 ^{\circ}C =273+28K=301K](https://tex.z-dn.net/?f=T_2%3D28%20%5E%7B%5Ccirc%7DC%20%3D273%2B28K%3D301K)
mole of gass will remain same at any emperature:
![\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)
putting all the value we get:
![T_2 =269.7K =-3.31 ^{\circ} C](https://tex.z-dn.net/?f=T_2%20%3D269.7K%20%3D-3.31%20%5E%7B%5Ccirc%7D%20C)
initial temperature=![-3.31^{\circ}C](https://tex.z-dn.net/?f=-3.31%5E%7B%5Ccirc%7DC)
Answer:
A. Antoine Lavoisier
Explanation:
He is know for creating the first chemical textbook. This textbook included rules for naming different chemical compounds, which is the still the same method we use to this day.
Hoped this can help! :D
sand silt and clay but also has large amounts of Carbonate. NOT carbonite.