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Professor
Steve
RossTo determine the number of atoms in 16g of oxygen gas, you first need to calculate the number of moles of oxygen using the molar mass of oxygen. Oxygen has a molar mass of 16g/mol. So, 16g of oxygen gas is equal to 1 mole of oxygen gas. One mole of oxygen gas contains approximately 6.022 x 10^23 atoms (Avogadro's number). Therefore, 16g of oxygen gas contains approximately 6.022 x 10^23 oxygen atoms.
If you know the density of oxygen and the molar mass of oxygen you can divide the two in order to find the molar volume of oxygen. (L/mol) Oxygen has a density of 1.429 g/L and a molar mass of 32.00 g/mol. (Just remember that the molar mass of oxygen, chemical formula of 0 2, comes straight from the Periodic Table.) So, divide: 1.429 g/L / 32.00 g/mol = 22.393 L/mol. The "g" unit cancells out and you're left with L/mol which is your molar mass. Compare that with 22.4 L/mol which is the accepted molar volume for an ideal gas, derived from the ideal gas law: pv=nRt. However, because most gases do not behave ideally there is a slight discrepancy.
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What is the mass in grams of 1mole of oxygen?
The mass of 2.000 mol of oxygen atoms is 32.00 grams.
How many grams of oxygen are contained in 18g of water?
The molar mass of water(H2O)=18((1*2)+16)The no. of moles(n) of water in 18 g of water=mass/molar mass=18 g/ 18 g mol-1 =1 molThe no. of molecules of water in 18 of water=n*Avogadro no. =1 mol*6.022*1023 mol-1 =6.022*1023The no. of atoms of oxygen in one mole of water=1 molThe no. of atoms of oxygen in 6.022*1023mol of water= 6.022*1023
How many oxygen atoms are present in 40.0 grams of oxygen gas?
There are approximately 0.625 moles of oxygen gas in 40.0 grams. Since the formula for oxygen gas is O2, there are 0.625 moles x 2 = 1.25 moles of oxygen atoms. Therefore, there are approximately 1.25 x 6.022 x 10^23 = 7.53 x 10^23 oxygen atoms in 40.0 grams of oxygen gas.
When 46g of sodium were exposed to dry air 62g of sodium oxide formed how many grams of oxygen from the air were used?
To find the amount of oxygen used, we need to consider the difference in mass between sodium and sodium oxide. The mass increase is 16g (62g - 46g) which corresponds to the amount of oxygen used from the air. Therefore, 16g of oxygen from the air were used.
How many grams of iron must be added to 16 grams of oxygen to equal 44 grams of iron 3 oxide?
Iron(III) oxide is composed of one iron atom and three oxygen atoms. Therefore, if we have 44 grams of iron(III) oxide, we can calculate the mass of iron by subtracting the mass of oxygen. 44g - 16g (mass of oxygen) = 28g of iron is needed to combine with 16g of oxygen to form 44g of iron(III) oxide.
