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What else can I help you with?
If you attach a manometer or pressure gauge to your two liter container what pressure would you measure?
The pressure measured in a closed two-liter container would depend on factors such as the temperature and the amount of gas or liquid inside the container. If the container is sealed and there is no chemical reaction occurring inside, the pressure would remain constant at the equilibrium pressure of the system.
What is the temperature of a 100 liter container having 1 mole of an ideal gas at a pressure of 20 kilo pascals?
To find the temperature, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for temperature (T), we have T = (PV) / (nR), where P is the pressure, V is the volume, n is the number of moles, and R is the ideal gas constant. Plugging in the given values (P = 20 kPa, V = 100 L, n = 1 mol, and R = 8.314 J/(mol·K)), we find that the temperature is approximately 239 K.
What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
If ambient room temperature is a constant 22 degrees celsius how much of a liter of water would evaporate in 12 hours and is the size of the exposed surface area of 1 liter relevant to evaporation?
-- the area of the exposed surface -- the airflow over the exposed surface -- the temperature of the liter of water -- the air pressure at the exposed surface -- the relative humidity of the air in the room -- the transparency of the liter container are all relevant to the rate of evaporation.
How many moles of O2 gas in 1L container?
I assume you mean at standard temperature and pressure. Use, PV = nRT (1 atm)(1 liter) = n(0.08206 L*atm/mol*K)(298.15 K) n = 1/24.466 = 0.04 moles oxygen gas --------------------------------
In a 2 liter container what is the partial pressure of oxygen?
The partial pressure of oxygen in a 2 liter container depends on the concentration of oxygen present in the container. If you know the concentration of oxygen in the container, you can use the ideal gas law to calculate the partial pressure. The formula is: partial pressure = concentration of oxygen x gas constant x temperature.
A 4.0 liter container of gas in initially at a pressure of 16 ATM what will the pressure be if the voume is changed to 1 liters at constant temperature?
Use Boyle's Law, applicable for ideal gases at constant temperature, to solve this problem: P1*V1 = P2*V2
Which element will uniformly fill a closed 2.0 liter container at STP?
At Standard Temperature and Pressure (STP), one mole of any ideal gas will occupy 22.4 liters. So to fill a 2.0 liter container at STP, you would need 2.0/22.4 = 0.089 moles of an ideal gas. This means any gas that is present in that amount and under those conditions can uniformly fill the container.
What will the pressure inside the container become if the piston is moved to the 1.80 L rm L mark while the temperature of the gas is kept constant?
This problem can be solved with the ideal gas law. The original pressure and volume of the container are proportional the final pressure and volume of the container. The original pressure was 1 atmosphere and the original volume was 1 liter. If the final volume is 1.8 liters, then the final pressure is 0.55 atmospheres.
a container of an ideal gas is at a constant volume what occurs when the volume of the contianer is doubled from 1 liter to 2 liters?
The volume doubles
If you were to transfer a liter of air into a 2-liter container how much of the 2-liter container would be filled?
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If you attach a manometer or pressure gauge to your two liter container what pressure would you measure?
The pressure measured in a closed two-liter container would depend on factors such as the temperature and the amount of gas or liquid inside the container. If the container is sealed and there is no chemical reaction occurring inside, the pressure would remain constant at the equilibrium pressure of the system.
Can you put ten liters of oxygen into one liter volume?
No, it is not possible to compress 10 liters of oxygen into a 1-liter volume. The volume of gas is dictated by its pressure and temperature through the ideal gas law, which means you cannot reduce 10 liters of gas into 1 liter without changing these properties significantly.
What is the temperature of a 100 liter container having 1 mole of an ideal gas at a pressure of 20 kilo pascals?
To find the temperature, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for temperature (T), we have T = (PV) / (nR), where P is the pressure, V is the volume, n is the number of moles, and R is the ideal gas constant. Plugging in the given values (P = 20 kPa, V = 100 L, n = 1 mol, and R = 8.314 J/(mol·K)), we find that the temperature is approximately 239 K.
What is the pressure of gas in a 30 liter container at 49 degrees c?
Gas pressure depends on volume, temperature, AND the amount of gas. You didn't give an amount of gas, so there is no way to answer your question.
What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
How do you measure 6 liter water from 7 liter and 5 liter jar?
There is most likely a more efficient way to do this, but this is the best I can do for now.Notation: ( x , y ) where x is the amount of water in the 5-liter container and y is the amount of water in the 7-liter container1. Fill the five-liter container ( 5 , 0 )2. Pour the five-liter container into the seven-liter container ( 0 , 5 )3. Fill the five-liter container ( 5 , 5 )4. Fill the seven-liter container with the five-liter container, leaving 3 liters in the five-liter container ( 3 , 7 )5. Pour out the seven-liter container ( 3 , 0 )6. Pour the five-liter container into the seven-liter container ( 0 , 3 )7. Fill the five-liter container ( 5 , 3 )8. Fill the seven-liter container with the five-liter container, leaving 1 liter in the five-liter container ( 1 , 7 )9. Pour out the seven-liter container ( 1 , 0 )10. Pour the five-liter container into the seven-liter container ( 0 , 1 )11. Fill the five-liter container ( 5 , 1 )12. Pour the five-liter container into the seven-liter container ( 0 , 6 )
