Answer:
Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
Explanation:
<u>Decomposition reaction:</u>
When a single compound break down into two or more simpler products.
For example "AB" reactant undergoes decomposition to form "A" and "B" products.
The chemical reaction is as follows.
The given compound is aluminium oxide.
The decomposition reaction of aluminium oxide is a follows.
The balanced equation is as follows.
Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
The balanced equation for the above reaction is as follows
P4 + 10Cl2 —> 4PCl5
P4 is the limiting reactant. Stoichiometry of P4 to PCl5 is 1:4
The mass of P4 reacted = 23.0 g
Number of P4 moles = 23.0 g/ 124 g/mol = 0.185 mol
According to the stoichiometry the number of PCl5 moles produced are 4 times the amount of P4 moles reacted
Therefore PCl5 moles = 0.185 x 4 = 0.74 mol
Answer:
0.01 faraday
Explanation:
We'll begin by writing the balanced equation. This is illustrated below:
CaCl₂ (aq) —> Ca²⁺ (aq) + 2Cl¯ (aq)
Ca²⁺ (aq) + 2e —> Ca
Molar mass of Ca = 40 g/mol
Mass of Ca from the balanced equation = 1 × 40 = 40 g
1 mole of electron (e) = 1 faraday
2 moles of electrons (e) = 2 × 1 faraday
2 moles of electrons (e) = 2 faraday
From the balanced equation above,
40 g of Ca was deposited by 2 faraday.
Therefore, 0.2 g of Ca will be deposited by = (0.2 × 2)/40 = 0.01 faraday.
Thus, 0.01 faraday is needed for the reaction.
