850,49 g of NF3 = 11,9784 moles
To determine the number of moles of fluorine present in 2.67 grams of nitrogen trifluoride (NF3), we first need to calculate the molar mass of NF3, which is 71.0 g/mol. Then we find the moles of NF3 in 2.67 g by dividing the mass by the molar mass: 2.67 g / 71.0 g/mol ≈ 0.038 moles. As there are three fluorine atoms in each molecule of NF3, there are 3 times more moles of fluorine: 0.038 moles NF3 x 3 = 0.114 moles of fluorine.
To convert grams to atoms, you need to first convert grams of nitrogen to moles using its molar mass (14.01 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, for 2.2 grams of nitrogen: Convert grams to moles: 2.2 g / 14.01 g/mol = 0.157 moles. Convert moles to atoms: 0.157 moles x 6.022 x 10^23 atoms/mol = 9.46 x 10^22 atoms of nitrogen.
To find the number of moles in 10.0 g of nitrogen dioxide (NO2), you first need to find the molar mass of NO2, which is approximately 46.01 g/mol. Then, divide the given mass by the molar mass to get the number of moles. In this case, 10.0 g / 46.01 g/mol ≈ 0.22 moles of nitrogen dioxide.
It is important to know that the percent of nitrogen in 4.444 moles of ammonium sulfide is the same as the percent of nitrogen in 454 grams or 4843 moles or 96 kg, etc. Remember the law of definite proportions - chemical compounds always contain the same proportion of elements by mass. Perhaps you were asking how much nitrogen is in 4.444 moles of ammonium sulfide given the percent of nitrogen in any given mass. So we'll do that too: find the percent of nitrogen in any given sample and apply it specifically to 4.444 moles.Before we go directly to the 4.444 moles, we have to figure out how much nitrogen is in any amount of ammonium sulfide by percent. To do this, we need the atomic weights of the elements and add them up to find the total molar mass of the compound.Ammonium sulfide = (NH4)2SNitrogen = 14.0 grams × 2 = 28.0 gramsHydrogen = 1.01 grams × 8 = 8.08 gramsSulfur = 32.1 grams------------------------------------------------------Ammonium sulfide = 68.2 gramsNow we take the mass of nitrogen and divide it by the total mass to get our percent.Nitrogen ÷ Ammonium sulfide = % Nitrogen28.0 grams ÷ 68.2 grams = 0.411 = 41.1% Nitrogen in Ammonium sulfideSince we know that in any amount of Ammonium sulfide contains 41.1% of Nitrogen, we can apply it to the mass given.41.1% of 4.444 moles = .411 × 4.444 = 1.83 moles of Nitrogen in 4.444 moles Ammonium sulfide
There are approximately 3.01 moles of urea in 25 g of CONH2, so there are 3.01 moles of nitrogen atoms. Therefore, there are 3.01 moles * 2 nitrogen atoms/molecule = 6.02 moles of nitrogen atoms in 25 g of CONH2, or urea.
The answer is 24,92 g nitrogen.
4.91 mol * 6.02214129(27)×1023 / mol = 2.96 ×1024
15 grams of nitrogen are equal to 1,071 moles.
There are 29/14, or just over 2 moles of nitrogen in 19 grams.
To determine the number of moles of fluorine present in 2.67 grams of nitrogen trifluoride (NF3), we first need to calculate the molar mass of NF3, which is 71.0 g/mol. Then we find the moles of NF3 in 2.67 g by dividing the mass by the molar mass: 2.67 g / 71.0 g/mol ≈ 0.038 moles. As there are three fluorine atoms in each molecule of NF3, there are 3 times more moles of fluorine: 0.038 moles NF3 x 3 = 0.114 moles of fluorine.
As a rule of thumb, the atomic mass of an element equals the number of grams of that element equals a mole. Since the atomic mass of Nitrogen is 14, there are 14 grams in one mole of Nitrogen. Next, we just have to divide 42 by 14 and we get our answer: There are 2.9988 moles in 42 grams of Nitrogen.
550 g of nitrogen dioxide is equal to 11,94 moles.
For this you need the atomic (molecular) mass of NO. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NO=30.0 grams6.64 grams NO / (30.0 grams)= .221 moles NO
the same amount would have to stay in grams, so if 14 grams of nitrogen is formed, then 8 grams of oxygen, add those two together and you get 22. and that's 22 of the 40 grams used, so 40 subtracted by 22 is 18. 18 grams of water would be formed.
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
To convert grams to atoms, you need to first convert grams of nitrogen to moles using its molar mass (14.01 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, for 2.2 grams of nitrogen: Convert grams to moles: 2.2 g / 14.01 g/mol = 0.157 moles. Convert moles to atoms: 0.157 moles x 6.022 x 10^23 atoms/mol = 9.46 x 10^22 atoms of nitrogen.
The clarity of this question is not clear. I will assume grams and possibly moles as that number does not look like a number of atoms. Grams first, then moles. 4.0 X 102 grams quinine (1 moleC20H2N2O2/302.236 grams)(2 moles N/1 mole C20H2N2O2) = 2.6 moles nitrogen =============== Pretty much the same procedure if you meant moles just no dividing out a mass.