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%
Answer:
Rhodium is used to make electrical contacts, as jewelry and in catalytic converters, but is most frequently used as an alloying agent in other materials, such as platinum and palladium. These alloys are used to make such things as furnace coils, electrodes for aircraft spark plugs and laboratory crucibles.
Explanation:
Answer: Theoretical yield of 
produced by 8.96 g of S is 33.6 g
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
2 moles of 
produce = 3 moles of
Thus 0.28 moles of will produce=
of
Mass of 
Thus theoretical yield of produced by 8.96 g of S is 33.6 g
Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1
Explanation:
