"<span>Changes of state are physical changes. They occur when matter absorbs or loses energy. Processes in which matter changes between liquid and solid states are freezing and </span>melting<span>. Processes in which matter changes between liquid and gaseous states are vaporization, evaporation, and condensation."</span>
Taking into account the reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Ca₃P₂:1 mole
- H₂O: 6 moles
- Ca(OH)₂: 3 moles
- PH₃: 2 moles
The molar mass of the compounds is:
- Ca₃P₂: 182 g/mole
- H₂O: 18 g/mole
- Ca(OH)₂: 74 g/mole
- PH₃: 34 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Ca₃P₂: 1 mole ×182 g/mole= 182 grams
- H₂O: 6 moles× 18 g/mole= 108 grams
- Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
- PH₃: 2 moles ×34 g/mole= 68 grams
<h3>Correct statements</h3>
Then, by reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
Answer:
Explanation:
We have the reactions:
A:
B:
Our <u>target reaction</u> is:
We have as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A:
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B:
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)