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1 mole of Ammonium Nitrate= 80g (R.A.M of compound)

Therefore 8g of Ammonium nitrate=

8g/80g= 0.1 mole

(moles = mass given over molar mass)

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The percent by mass of nitrogen in NH4NO3 is closest to?

Mass of Nitrogen: 14g/mol Mass of Ammonium Nitrate (NH4NO3): 14 + 1x4 + 14 + 16x3 = 80g/mol ∴ % of Nitrogen in Ammonium Nitrate = 14/80 * 100 = 17.5%


What is the percent by mass of oxygen in NO2?

first we work out the Mr. Mr= 80g Then we find out the Mr of Oxygen in NH4NO3. Mr of oxygen=48g Then we do 40/80 which equals 0.5 then we multiply it by 100 and we get 50% This is our Percentage by mass.


How many grams of KCl remain in solution at 20 degrees celsius when a solution containing 80g of KCl in 200g of water at 50 degrees Celsius is cooled to 20 degrees Celsius?

To find the amount of KCl that remains in solution at 20 degrees Celsius, you can use the principle of solubility. Calculate the maximum amount of KCl that can dissolve in 200g of water at 20 degrees Celsius using a solubility chart. Once you have this value, compare it to the initial 80g of KCl to determine how much remains in solution after cooling.


When 80g of sodium hydroxide are dissolved in water to make 0.500 of a solution what is the molarity?

The molarity can be calculated by first converting grams of sodium hydroxide to moles, then dividing by the volume of the solution in liters. In this case, the molar mass of NaOH is 40 g/mol. So, 80g is 2 moles. With a volume of 0.5 L, the molarity is 2 moles / 0.500 L = 4 M.


How many grams of sodium does a 5.35 g sample of sodium bromide contain?

All you really need here is a ratio of molar masses, ie molar mass of Na/molar mass of NaBr So look at your periodic table and is says: Na = 23g/mol Br = 80g/mol therefore NaBr = 103g/mol so from this, you actually know a percentage of Sodium to Bromine, correct me if I'm wrong, someone else please but just take: (23g/mol)/103g/mol = 22.3% of you NaBr is actually Na, so take 5.35g * 0.223 and you get: 1.19g of Na in your NaBr Hope that helps. Cheers