Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.

Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state

And copper goes from +2 to 0

On balancing the number of electrons we get:
For 1 mole of Al
is required.
Therefore for 2 moles of Al,
Total
F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
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Boiling point
i hope this helps.
Answer:
0.162 moles of CO₂ are produced by this reaction
Explanation:
The reaction is:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)
As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.
Density = Mass / Volume → Density . Volume = Mass
0.00183 g/mL . 1300 mL = Mass → 2.379 g
We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide
Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles
Answer:
They all wear their skeletons on the outside! This is called an exoskeleton and all creatures that have their structure on the outside are included in the phylum of Arthropods.
Explanation:
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)