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When a gas of 3.4 moles occupies a volume of 40.6 L at 298 K what is the pressure of the gas in kPa?
Wiki User
∙ 6y ago
The pressure is 207,5 kPa.
Wiki User
∙ 6y ago
Sarah B
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How many moles in 5.67 grams of fluorine gas?
.298 Moles
What woul happen to the volume of the 100L weather balloon if its temperature increases from 25 C TO 125 C while the external pressure remain constant?
The universal gas equation is PV = nRT (Pressure x Volume = Number of moles x Universal Gas Constant x Temperature in Kelvin/Rankin). So - if Pressure is constant, the number of moles is constant, but the temperature increases from 25C (298 K) to 125C (398K) - a 34% increase, a similar 34% increase in volume will occur.
A sample of a gas pressure of 3.00atm at 25c what would the gas pressure be at 52c if the volume remaiins constant?
25C is 298K. 52C is 325K. Assuming linearity (an ideal gas), P=3*325/298=3.27 atm.
What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?
Molar Mass O2: 32.00, Molar Mass CO2: 44.01, T= 21C+273= 294Kelvins V=5.00 L Grams of O2=20.0 so moles O2= 20.0g/32.00g=.625g/mol Grams of CO2=4.4 so moles CO2= 4.4g/ 44.01=.1 Total moles: .625+.1= .725 mols (this is n) Ideal Gas Law: PV=nRT....Rearrange and solve for pressure. P=nRT/V.....R is the constant which is equal to .08206L*atm*mol P=(.725)(.08206)(294K)/(5.00L) P= 3.5atm
How many moles are contained in 22.4 L N2 at STP?
This is the molar volume atstandard pressure po= 1 ATM (= 101.325 kPa)and atstandard temperature (melting ice) T= 273.15 K(At room temperature T= 298 K and p= po the molar volume is 24.5 L/mole)
How many moles in 5.67 grams of fluorine gas?
.298 Moles
What woul happen to the volume of the 100L weather balloon if its temperature increases from 25 C TO 125 C while the external pressure remain constant?
The universal gas equation is PV = nRT (Pressure x Volume = Number of moles x Universal Gas Constant x Temperature in Kelvin/Rankin). So - if Pressure is constant, the number of moles is constant, but the temperature increases from 25C (298 K) to 125C (398K) - a 34% increase, a similar 34% increase in volume will occur.
How many pages is Pride and Prejudice?
In my copy of Pride and Prejudice, the novel itself occupies 298 pages.
What is the pressure of a can with a volume of 250 mL containing 2.3 g of C3H8 if the can is at 298 K?
You must use the equation PV = nRT, where P = pressure in ATM V = volumes in L n = moles of gas R = gas constant (.0821 mol*ATM/L*K) T = Temperature in K 2.3g C3H8 / 44.09 g (molecular weight) = 0.052 moles 250mL = 0.25 L P(0.25 L) = (0.052 moles)(0.0821 mol*ATM/L*K)(298K) P = 5.09 ATM
0.783 g of hydrogen gas is placed in an container of volume 657 mL at a temperature of 342 K What is the pressure in ATM?
0.783 g of hydrogen gas is placed in an container of volume 657 mL at a temperature of 342 K. What is the pressure in atm?
What page was mr. darcy's proposal in Pride and Prejudice?
In my volume of Pride and Prejudice, Darcy's first, unsuccessful, proposal begins on page 145, of the 298 pages in the novel. It will be different in different editions, but clearly this is only slightly less than half way through. It takes up almost the whole of chapter XI of volume II. Another edition I have has it on page 113 of 236, and the chapter is listed as chapter 34.
A sample of a gas pressure of 3.00atm at 25c what would the gas pressure be at 52c if the volume remaiins constant?
25C is 298K. 52C is 325K. Assuming linearity (an ideal gas), P=3*325/298=3.27 atm.
What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?
Molar Mass O2: 32.00, Molar Mass CO2: 44.01, T= 21C+273= 294Kelvins V=5.00 L Grams of O2=20.0 so moles O2= 20.0g/32.00g=.625g/mol Grams of CO2=4.4 so moles CO2= 4.4g/ 44.01=.1 Total moles: .625+.1= .725 mols (this is n) Ideal Gas Law: PV=nRT....Rearrange and solve for pressure. P=nRT/V.....R is the constant which is equal to .08206L*atm*mol P=(.725)(.08206)(294K)/(5.00L) P= 3.5atm
How many charts are in 298 gallons?
There is no such thing as "charts" in volume. It may be possible that you mean 'quarts', in which case there are 1,192 quarts in 298 gallons.
A sample of gas occupies 2.78 x 103 mL at 25oC and 760 mm Hg What volume will the gas sample occupy at the same temperature and 475 mm Hg?
The volume that the gas sample will occupy at the same temperature and 475 mmHg is 4448 mL. In significant figures, the answer would be 4400 mL To find this, you can use the Combined Gas Law, (P1V1)/T1=(P2V2)/T2. First, you need to convert your temperature from degrees Celsius to Kelvin. You can do this by adding 273 to 25, which gives you 298 K. Then you can plug in the given values for volume, pressure, and temperature. The equation should look like this: (760 mmHg * 2.78 x 103 ml) / 298 K = (475 mmHg * V2) / 298 K Then you can solve for V2 to find the unknown volume.
How many moles are contained in 22.4 L N2 at STP?
This is the molar volume atstandard pressure po= 1 ATM (= 101.325 kPa)and atstandard temperature (melting ice) T= 273.15 K(At room temperature T= 298 K and p= po the molar volume is 24.5 L/mole)
What is the pressure in ATM exerted by 25.0 grams of XeF6 in a 5 liter container at 25 degrees Celsius?
Use the Ideal Gas Law: PV=nRT (pressure times volume equals moles times the ideal gas constant of 0.082 times the Kelvin temperature.) First, convert 25 Celsius to Kelvin: 25+273=298. Next, find out how many moles is in 25.0g of XeF6: 1(Xe)+6(F)=1(131.3)+6(19)=245g/mol; 25g XeF6=0.102mol. Then, plug it all in and solve for the pressure, which is your unknown: P(5L)=0.102mol(0.082)(298kelvins) Simplify: 5P=2.4925, P=0.4985, or approximately 0.5 atm of pressure.
