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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?
3.0 atm
What is the density of Cl2 gas at STP?
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
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
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?
3.0 atm
A 8.58 cylinder of O2 gas contains 89.6 g O2 at a temperature of 21 celsius what is the pressure?
7.88 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 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 )
What is the density of Cl2 gas at STP?
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
How many moles of oxygen (O2) are present in 33.6 L of the gas at 1 ATM and 0 and deg C?
The number of moles is 1,5.
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
Calculate the partial pressures at equillibrium?
more info::Suppose 1.20 ATM of CH4(g), 2.03 ATM of C2H6(g), and 15.69 ATM of O2(g) are placed in a flask at a given temperature. The reactions are given below.CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) KP = 1.0 1042 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) KP = 1.0 108Calculate the equilibrium pressures of all gases.PCH4Enter a number with the correct number of significant figures.1atmPC2H6Enter a number with the correct number of significant figures.2atm
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
How many liters of oxygen are necessary for the combustion of 277 g of carbon monoxide assuming that the reaction occurs at stp the balanced equation is 2 co plus o2 equals 2 co2?
2CO + O2 -> 2CO2 find moles O2 gas 277 grams carbon monoxide (1 mole CO/28.01 grams)(1 mole O2/2 mole CO) 4.9447 moles O2 Now, since I am lazy, I use PV = nRT V = nRT/P V = (4.9447 moles O2)(0.08206 L*atm/mol*K)(298.15 K)/(1 atm) V = 120.978 Liters of oxygen, or, in sigi figi; 121 Liters of oxygen needed for this combustion
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
