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What is the volume of oxygen occupied by two moles?
The molar volume of a gas at STP (standard temperature and pressure) is 22.4 L/mol. Therefore, the volume occupied by 2 moles of oxygen would be 44.8 L.
What is the volume of oxygen that will be used by 12g of Mg to be completely converted into MgO?
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
Does oxygen and nitrogen weigh the same in volume?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
How many grams are contained in 0.644 mol oxygen?
To find the grams in 0.644 mol of oxygen, you need to multiply the number of moles by the molar mass of oxygen. The molar mass of oxygen is approximately 16 g/mol. Therefore, 0.644 mol of oxygen would contain 0.644 mol x 16 g/mol = 10.304 grams of oxygen.
How many grams of oxygen gas can be formed from the decomposition of 0.437 mol HgO?
By using the balanced chemical equation for the decomposition of mercury(II) oxide (HgO): 2 HgO -> 2 Hg + O2, we see that 1 mol of HgO produces 1 mol of O2. Therefore, 0.437 mol of HgO will produce 0.437 mol of O2. To convert mol to grams, we use the molar mass of oxygen: 32.00 g/mol. so, 0.437 mol of O2 is equivalent to 0.437 mol * 32.00 g/mol = 13.92 grams of O2.
What is the volume of oxygen occupied by two moles?
The molar volume of a gas at STP (standard temperature and pressure) is 22.4 L/mol. Therefore, the volume occupied by 2 moles of oxygen would be 44.8 L.
What is the volume of oxygen that will be used by 12g of Mg to be completely converted into MgO?
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
How do you figure out the HNO3 Concentration?
The equation you will need is: Mol of substance 1 * volume of substance 1 = Mol of substance 2 * volume of substance 2
What volume of oxygen is needed to react with solid sulfur to form 6.20L of sulfur dioxide?
1 mol of sulfur reacts with 1 mol of oxygen to form 1 mol of sulfur dioxide. The balanced chemical equation is: S(s) + O2(g) -> SO2(g). Since gases occupy the same volume under the same conditions, the volume of oxygen needed would also be 6.20L.
What volume of oxygen is needed to react with a solid sulfur to form 3.5L of SO2?
1 mole of gas occupies 22.4 liters at STP. Therefore 3.5/22.4 = 0.15625 moles of SO2. There are thus 0.15625 moles of O2 needed to react with solid sulfur because S + O2 ---->SO2. 0.15625 moles of oxygen occupies 0.15625 x 22.4 liters = 3.5 liters O2 required.
A mixture of gases contains 0.25 mol oxygen gas 0.75 mol nitrogen gas and 1.50 mol carbon dioxide gas at STP The mole fraction of oxygen gas in this mixture is?
At STP (Standard Temperature and Pressure), the volume occupied by 1 mol of any gas is 22.4 L. Thus, 0.25 mol of oxygen gas occupies (0.25 mol) * (22.4 L/mol) = 5.6 L in the mixture. The total volume of the gas mixture can be found as the sum of the individual volumes of oxygen, nitrogen, and carbon dioxide. Then, the mole fraction of oxygen gas is the moles of oxygen gas divided by the total moles of all gases in the mixture.
Which gas is heavier oxygen or nitrogen?
Since both form diatomic elements, we simply have to compare molecular masses. O2 has a mass 32.0 g/mol, while N2 is 28.0 g/mol. This means that, since gases (according to the ideal gas law) all contain 22.4 mol/L, then the same volume of oxygen would be heavier than the same volume of nitrogen.
Does oxygen and nitrogen weigh the same in volume?
No, one mole of each, having the same VOLUME (about 22.4 L at STP), differ though in their masses: 32 g/mol for O2 and 28 g/mol for N2 So their densities (mass per volume) also differ in the same way: 1.43 g/L and 1.25 g/L respectively, at STP.
Calculate the number of moles of oxygen that will dissolve in 45 L of water at 20C if the partial pressure of oxygen is 0.21 ATM Henrys constant for oxygen is 0.0013 mol L ATM?
Using Henry's law, the number of moles of oxygen that will dissolve is calculated by multiplying Henry's constant by the partial pressure of oxygen and the volume of water. So, moles of oxygen = 0.0013 mol L ATM * 0.21 ATM * 45 L = 0.1233 mol of oxygen will dissolve in 45 L of water at 20C.
What is mass and no of molecules of O2 having volume 22.4 dm3?
22,4 L is the molar volume. 1 mol of oxygen has 32 g and and 6,022140857.10e23 molecules.
How many grams are contained in 0.644 mol oxygen?
To find the grams in 0.644 mol of oxygen, you need to multiply the number of moles by the molar mass of oxygen. The molar mass of oxygen is approximately 16 g/mol. Therefore, 0.644 mol of oxygen would contain 0.644 mol x 16 g/mol = 10.304 grams of oxygen.
How many grams of oxygen gas can be formed from the decomposition of 0.437 mol HgO?
By using the balanced chemical equation for the decomposition of mercury(II) oxide (HgO): 2 HgO -> 2 Hg + O2, we see that 1 mol of HgO produces 1 mol of O2. Therefore, 0.437 mol of HgO will produce 0.437 mol of O2. To convert mol to grams, we use the molar mass of oxygen: 32.00 g/mol. so, 0.437 mol of O2 is equivalent to 0.437 mol * 32.00 g/mol = 13.92 grams of O2.
