Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
I believe the answer is D
Element 87 francium have the highest atomic size, so in short the closest element to francium have the high atomic size.
Answer:
0.14 moles of Fe₂O₃
Explanation:
Given parameters:
Number of moles of Fe = 0.27moles
Unknown:
Number of moles of Fe₂O₃ = ?
Solution:
To solve this problem, we are going to the work from the known specie to the unknown using their number of moles.
We first obtain the balanced equation of the reaction;
4Fe + 3O₂ → 2Fe₂O₃
The equation above is balanced;
4 moles of Fe produced 2 moles of Fe₂O₃
0.27 moles of Fe will produce
= 0.14 moles of Fe₂O₃
Answer:
<h3>The answer is 8.98 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of cube = 150 g
volume = 16.7 cm³
We have

We have the final answer as
<h3>8.98 g/cm³</h3>
Hope this helps you