Answer:
0,13 moles of acid are produced
Explanation:
The reaction of the problem is:
PCl₅ + 4H₂O → 5HCl + H₃PO₄
Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).
The molecular mass of PCl₅ is:
1P = 30,97g/mol + 5Cl = 5×35,45g/mol = <em>208,24 g/mol</em>
That means 4,5g of PCl₅ are:
4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:
0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =
<em>0,13 moles of acid are produced</em>
I hope it helps!
Answer:
answer is C
Explanation:
encourage the release of carbon dioxide from the stems
The answer is 2H2 + O2----> 2H2O
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
1. Dmitri Mendeleev
2. Johann Dobhereiner
3. John Newlands
4. Henry Moseley
If you find this helpful please give me Brainliest award