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A tank contains N2 at 1.0 ATM and O2 at 2.0 ATM Helium is added to this tank until the total pressure is 6.0 ATM What is the partial pressure of the helium?
The initial total pressure is 1.0 ATM + 2.0 ATM = 3.0 ATM. Therefore, 6.0 ATM - 3.0 ATM = 3.0 ATM of helium was added to the tank. Hence, the partial pressure of helium in the tank is 3.0 ATM.
What is the partial pressure of O2g in a mixture of gases consisting of 2.50 g of O2 and 6.40 g of Heg if the total pressure of this mixture is 8.25 ATM?
Moles of O2=2.5/32=0.078125 Moles of He=6.4/4=1.6 Mole fraction of O2=0.078125/(0.078125+1.6)=0.046555 Partial pressure of O2=0.046555 x 8.25=0.384atm
How many grams of O2 are in a 351 ml container that has a pressure of 12 atm at 25 Celsius?
You will be using the equation for the ideal gas law: PV=nRT, where P=pressure in atm, V=volume in liters, n=moles, R=ideal gas constant (0.0821) and T=temperature in kelvins, and along the way, you must convert mL to L and Celsius to kelvins. So, plug it all in: 12atm(0.351L) = n(0.0821)(298k). The answer is 0.17mol O2, which is about 2.7g O2.
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
What is the volume of 5.60 g of O2 at 7.78 ATM and 415 K?
To find the volume of 5.60 g of O2 at 7.78 ATM and 415 K, we can use the Ideal Gas Law, ( PV = nRT ). First, we need to calculate the number of moles of O2: ( n = \frac{5.60 , \text{g}}{32.00 , \text{g/mol}} = 0.175 , \text{mol} ). Using the Ideal Gas Law rearranged to ( V = \frac{nRT}{P} ), and substituting ( R = 0.0821 , \text{L·atm/(K·mol)} ), we get: [ V = \frac{(0.175 , \text{mol})(0.0821 , \text{L·atm/(K·mol)})(415 , \text{K})}{7.78 , \text{ATM}} \approx 1.03 , \text{L}. ] Thus, the volume of 5.60 g of O2 at the given conditions is approximately 1.03 liters.
What is the density of o2 at 1 ATM and 20 degrees celsius?
The density of O2 at 1 atm and 20 degrees Celsius is approximately 1.429 g/L.
A tank contains N2 at 1.0 ATM and O2 at 2.0 ATM Helium is added to this tank until the total pressure is 6.0 ATM What is the partial pressure of the helium?
The initial total pressure is 1.0 ATM + 2.0 ATM = 3.0 ATM. Therefore, 6.0 ATM - 3.0 ATM = 3.0 ATM of helium was added to the tank. Hence, the partial pressure of helium in the tank is 3.0 ATM.
How many grams of KO2 would be required to produce 1120.0L of O2 at 20.0 degrees Celsius and 1.00atm?
Use.PV = nRTTo get moles O2. ( 20.0o C = 293.15 Kelvin )(1.00 atm)(1120.0 L) = (X moles)(0.08206 L*atm/mol*K)(293.15 K)Moles = 1120.0/24.56= 45.60 moles O2-----------------------------------now,45.60 moles O2 (1 mole KO2/2 mole O2)(71.1 grams/1 mole KO2)= 1621 grams potassium oxide required=============================
How many moles of oxygen (O2) are present in 33.6 L of the gas at 1 ATM and 0 and deg C?
The ideal gas law, PV = nRT, can be rearranged to solve for moles: n = PV / RT. At 1 atm and 0°C (273 K), R = 0.0821 Latm/molK. Plugging in the values: n = (1 atm * 33.6 L) / (0.0821 Latm/molK * 273 K) = approximately 1.36 moles of O2.
What is the partial pressure of O2g in a mixture of gases consisting of 2.50 g of O2 and 6.40 g of Heg if the total pressure of this mixture is 8.25 ATM?
Moles of O2=2.5/32=0.078125 Moles of He=6.4/4=1.6 Mole fraction of O2=0.078125/(0.078125+1.6)=0.046555 Partial pressure of O2=0.046555 x 8.25=0.384atm
How many grams of O2 are in a 351 ml container that has a pressure of 12 atm at 25 Celsius?
You will be using the equation for the ideal gas law: PV=nRT, where P=pressure in atm, V=volume in liters, n=moles, R=ideal gas constant (0.0821) and T=temperature in kelvins, and along the way, you must convert mL to L and Celsius to kelvins. So, plug it all in: 12atm(0.351L) = n(0.0821)(298k). The answer is 0.17mol O2, which is about 2.7g O2.
Find the mass of oxygen in a cylinder having a capacity of 100 liters at a temperature of 20 degree Celsius The pressure in the cylinder is maintained at 80 bar R equals 0.259859?
Let us do some conversions first, then use the ideal gas law. PV = nRT 1 Bar = 1.01325 atmospheres 80 Bar (1.01325 ATM/1 Bar) = 81.06 ATM 20 C = 293.15 K new R = 0.08206 L*ATM/mol*K (81.06 ATM)(100 L) = nI0.08206 L*ATM/mol*K)(293.15 K) moles O2 = 337 moles O2 * 32 grams = 10784 grams about 24 pounds of oxygen
IS o2 an ideal gas at 35 ATM and 33.15K?
It isn't even a gas at that temperature let alone being ideal. The boiling point of O2 is 90 K and the melting point is 55K. So at 33 K it is a solid. At extremely low pressures this might change a bit, but you gave a huge pressure so not the case.
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
What is the volume of 5.60 g of O2 at 7.78 ATM and 415 K?
To find the volume of 5.60 g of O2 at 7.78 ATM and 415 K, we can use the Ideal Gas Law, ( PV = nRT ). First, we need to calculate the number of moles of O2: ( n = \frac{5.60 , \text{g}}{32.00 , \text{g/mol}} = 0.175 , \text{mol} ). Using the Ideal Gas Law rearranged to ( V = \frac{nRT}{P} ), and substituting ( R = 0.0821 , \text{L·atm/(K·mol)} ), we get: [ V = \frac{(0.175 , \text{mol})(0.0821 , \text{L·atm/(K·mol)})(415 , \text{K})}{7.78 , \text{ATM}} \approx 1.03 , \text{L}. ] Thus, the volume of 5.60 g of O2 at the given conditions is approximately 1.03 liters.
How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?
I assume you mean the decomposition reaction used to produce O2 in lab. 2KCLO3 -> 2KCl + 3O2 find moles O2 55.2 grams KClO3 (1 mole KCLO3/122.55 grams)(3 mole O2/2 mole KClO3) = 0.67564 moles O2 Now, I use the ideal gas law PV = nRT (1 atm)(V) = (0.67564 mol)(0.08206 L*atm/mol*K)(298.15 K) Volume O2 = 16.53 Liters which is 16530 milliliters ( less significant figures )
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 --------------------------------
