The amount of atoms of Phosphorus in 3.80 mol of copper (II) phosphate would be equivalent with :
3.80 mol x (6.022 x 10^23 molecules)/ (1 mol)
Hope this helps
Answer:
Al(s) ⇄ Al³⁺(aq) + 3e⁻ (oxidation)
Mg⁺²(aq) + 2e⁻ ⇄ Mg(s) (reduction)
Explanation:
In the cell notation, first is placed the anode and then the cathode, so Al(s) is in the anode, where an oxidation reaction is taking place, and Mg²⁺ is in the cathode, where a reduction reaction is taking place. The half-reactions are:
Al(s) ⇄ Al³⁺(aq) + 3e⁻ (oxidation)
Mg⁺²(aq) + 2e⁻ ⇄ Mg(s) (reduction)
For the global reaction, it'll be necessary that the number of electrons (e⁻) is equal in the half-reactions, so the first one is multiplied by 2 and the second by three and the global reaction is:
2Al(s) + 3Mg⁺²(aq) ⇄ 2Al³⁺(aq) + 3Mg(s)
Answer:
51:49 is the ratio of female participants to male participants.
Explanation:
Percentage of females in the race = 51 %
Percentage of males in the race = 100% - 51 % = 49%
Total participants in race = 400
Number of women participants = 
Number of men participants = 
Ratio of female participants to male participants :
Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
