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What is the volume of the gas at 200 degrees c if it is 25 L at 0 degrees c?
Use Charles's Law: V1 / T1 = V2 / T2 Constant pressure must be kept. Absolute temp. must be used. T1 = 0 + 273 = 273 degr.K T2 = 200 + 273 = 473 degr.K 25 L / 273 K = V2 / 473 K V2 = 25 x 473 / 273 = 43.315 L (final volume).
A sample of air has a volume of 140ml at 67degrees Celsius At what temperature will its volume be 50ml at constant pressure?
V1/T1 = V2/T2 Where temperature must be in Kelvins 67C + 273 = 340 K So 140/340 = 50/T2 Find T2 340/140(50) = T2 T2 = 121 K or -152C
Chemisrty if there are 288 degrees c how many is converted to k degrees?
288 plus 273 equals 561 K. (Any)degrees C +273 = degrees K
What volume is needed to store 10kg of o2 at 7.5 ATM pressure and 294k?
(PV / nT )a = (PV / nT )b where the a and represent different conditions At STP, 1 mole of a gas occupies 22.4 L --- these are conditions b conditions a are the 7.5 atm, 294 K, but the 10 Kg must be converted to moles (n). 1 mole O2 = 32 g, so 10 kg x (1000 g / 1 kg) x (1 mole / 32 g) = 312.5 moles (7.5 atm)x(V) / [(312.5 moles)(294 K)] = (1 atm)(22.4L)/[(1 mole)(273 K)] Rearrange and solve for V V = (1 atm)(22.4L)/[(1 mole)(273 K)] x (312.5 moles)(294 K)/(7.5 atm) V = 1005.1 Liters
Zinc normal phase?
Solid under standard pressure and temperature conditions (0 degree Celsius, 273 K, and 1 atm)
What is the volumeof one mole of oxygen gas at 273 K and 760 torr?
The volume of one mole of any ideal gas at standard temperature and pressure (273 K and 1 atm) is approximately 22.4 L. Since the pressure given is 760 torr, which is equivalent to 1 atm, the volume of one mole of oxygen gas at 273 K and 760 torr would also be approximately 22.4 L.
What is the volume occupied by 2.20 mol of argon at STP?
The volume occupied by 2.20 mol of argon at standard temperature and pressure (STP) is approximately 49.68 liters. This is calculated using the ideal gas law equation, V = nRT/P, where n is the number of moles, R is the gas constant, T is the temperature, and P is the pressure. At STP, T = 273 K and P = 1 atm.
What is the volume of the gas at 200 degrees c if it is 25 L at 0 degrees c?
Use Charles's Law: V1 / T1 = V2 / T2 Constant pressure must be kept. Absolute temp. must be used. T1 = 0 + 273 = 273 degr.K T2 = 200 + 273 = 473 degr.K 25 L / 273 K = V2 / 473 K V2 = 25 x 473 / 273 = 43.315 L (final volume).
If a round balloon has a radius of 10.0 cm at 273 K what will the radius be at 373 K?
The volume will be 5723 assuming that the balloon's elasticity makes no difference.
Which changes in pressure and temperature occur as a given mass of gas at 380 torr and 546 K is changed to STP known?
To change from 380 torr and 546 K to standard temperature and pressure (STP: 1 atm and 273 K), the pressure decreases from 380 torr to 1 atm, and the temperature decreases from 546 K to 273 K. This change can be calculated using the ideal gas law and the principles of gas behavior.
Which shows the formula for converting from kelvins to degrees Celsius C (95 K) plus 32 C 59 (K 32) C K 273 C K plus 273?
The correct formula for converting from kelvins to degrees Celsius is C = K - 273. So, to convert 95 K to degrees Celsius, you would do C = 95 - 273 = -178 degrees Celsius.
What is the formula of celsius to kelvin?
K= C + 273 Example- the freezing point of water is 0 c 0 + 273= 273. So water freezes at 273 k
What volume of air contains 12.7g of oxygen gas at 273 K and 1.00 ATM?
Using the ideal gas law, PV = nRT, we can calculate the moles of oxygen gas present. Then, using the molar mass of oxygen, we can find the volume of air required to contain 12.7g of oxygen gas at 273 K and 1.00 atm.
How many moles of oxygen must be placed in a container with a volume of 50 liters to produce pressure of 4.5 ATM at standard temperature?
Using the ideal gas law (PV = nRT), we can rearrange the equation to solve for moles: n = PV / RT. Given the pressure (P = 4.5 ATM), volume (V = 50 L), standard temperature (T = 273 K), and the gas constant (R = 0.0821 L · ATM/mol · K), we can calculate the number of moles of oxygen needed. n = (4.5 ATM * 50 L) / (0.0821 L · ATM/mol · K * 273 K) ≈ 8.38 moles.
0C -273 K True?
273
How do you find the volume if you have the pressure and temperature?
Use general gas law: V = n.R.T / pin which:n = number of moles (to be filled in)R = gas constant = 8.20*10-2 (L.atm.K-1.mol-1)T = tempeature (K) = 273 K (stand.T)p = pressure (atm) = 1.00 atm (stand.P)then the calculated volume is in Liter
How many liters of radon gas would be in 3.43 moles at room temperature and pressure (273 K and 100 kPa)?
There are 22.4 Liters per mole at standard temperature and pressure. (273 K and 1 atm). So 100 kPa = 0.9869 atm, so 1 mole occupies 22.1066 L. Multiply by 3.43 mole * (22.1055 L/mole) = 75.826 L. Note that I would not consider 273 room temperature. It is the same as 0°C or 32°F. Usually they say 293 K is room temperature.
