Answer:
because a fruit holds the seeds it needs to reproduce and continue living.
Answer:
332.918g O2
Explanation:
I'm having some issues with the work however, your final answer should be 332.918g O2
Hope this helped!
Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
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:
Calcium chloride react with aluminum to produce calcium and aluminum chloride. This reaction takes place at a temperature of 600-700°C.
CaCl2+F2 -> CaF2 + Cl2