The answer is 4.69 x 10⁻¹⁹ I hope this helped!
Fe3N2, also known as Iron (II) nitride, is an ionic compound.
Ionic compounds are compounds that consists of metals and non-metals bonded with ionic bonds. The metal ion gives up electron(s) to the non-metals.
Since iron is a metal and nitrogen is an non-metal, the bond they would form would be an ionic bond. Iron gives up 2 electrons to form iron(II) ion, while nitrogen gains 3 electrons to form nitride ion. Since one iron cannot let a nitrogen gain 3 electrons, so in the compound, there would be 3 iron (ii) ions that has given up 6 electrons in total while 2 nitride ions have gained 6 electrons in total.
MgF2 is <span>Magnesium fluoride
hope this helps!</span>
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
