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:
6 x 10⁶ g Fe
Explanation:
Step 1: Set up dimensional analysis
7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)
Step 2: Multiply, divide, and cancel out units
atoms Fe and atoms Fe cancel out.
mol Fe and mol Fe cancel out.
We should be left with g Fe.
7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe
116279(55.85) = 6.49 x 10⁶ g Fe
Step 3: Sig figs
There is only 1 sig fig in this problem.
6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe
<u>Answer:</u> The original element is 
<u>Explanation:</u>
Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.
The released alpha particle is also known as helium nucleus.
For the given alpha decay process of an isotope:
<u>To calculate A:</u>
Total mass on reactant side = total mass on product side
A = 208 + 4
A = 212
<u>To calculate Z:</u>
Total atomic number on reactant side = total atomic number on product side
Z = 82 + 2
Z = 84
The isotopic symbol of unknown element is
Hence, the original element is
Answer:
2 half-lives=0.8
6 half-lives= 0.05
Explanation
Half-lives are constant and always decrease by half, implying that the concentration decreases by half at a consistent rate.
3.2/2= 1.6/2= 0.8 is two half-lives
3.2/2= 1.6/2= 0.8/2= 0.4/2= 0.2/2= 0.1/2=0.05 is six half-lives
