Assuming that this ammonia gas is at STP, you can use Avogadro's number to gind the number of moles of gas:
(387 x 1021 molecules) x (1 mol / 6.02x1023particles) x (17.03 g / 1 mol) =110 g NH3
394.794 grams
The answer is 20,664 g ammonia.
1.60 x 10^24 molecules
89,6 g ammonia are obtained.
To calculate the number of grams in 4.1 x 10^22 molecules of N2I6, you first need to find the molar mass of N2I6. Then, use this molar mass to convert the number of molecules to grams using Avogadro's number and the formula: grams = (number of molecules) / (Avogadro's number) * molar mass.
To produce 525 grams of ammonia (NH3), you would need 25 moles of ammonia. Since the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is 3H2 + N2 -> 2NH3, you would need 75 moles of hydrogen molecules (H2) to produce 525 grams of ammonia. This is equivalent to 4,500 molecules of hydrogen.
There are 200 grams of ammonia in 200 grams of ammonia.
This value is 21,85 g.
There are 6.02x10^23 molecules in one mole of anything.
394.794 grams
0,522 moles of ammonia contain 3,143.10e23 molecules of NH3.
To find the number of moles in 170000 grams of ammonia, you need to divide the given mass by the molar mass of ammonia. The molar mass of ammonia (NH3) is about 17 grams/mol. Therefore, 170000g ÷ 17g/mol ≈ 10000 moles of ammonia.
3 moles of ammonia is 51grams. One mole is 17 grams.
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
7,02 g ammonia
The answer is 20,664 g ammonia.
To find the number of moles in 1200 grams of ammonia, divide the given mass by the molar mass of ammonia. The molar mass of ammonia (NH3) is approximately 17 grams/mole. Therefore, 1200 grams divided by 17 grams/mole equals approximately 70.59 moles of ammonia.