Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
Answer:
63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.
Explanation:
I this is college level chemistry not regular high school chem.
A moon would be the correct answer.
Answer:
Four times the original amount if only one orange was used
Explanation:
We can assume that the oranges all have equal voltages. Connecting them in series will have an increasing effect on the voltage delivered. In our case, this will produce 4 times the voltage of the circuit when only one orange is used.
Whenever simple cells are connected in series, the voltages of the individual cells are added up to form the voltage of the whole circuit.
Let us assume that the voltage of each of the oranges is approximately 0.9 volts. The Voltage produced when the 4 oranges are joined in series is 0.9 + 0.9 + 0.9 + 0.9 = 3.6 volts
I believe the answer would be true