Answer:
The molar mass of the unknown acid is 386.8 g/mol
Explanation:
Step 1: Data given
Mass of the weak acid = 1.168 grams
volume of NaOH = 28.75 mL = 0.02875 L
Molarity of NaOH = 0.105 M
Since we only know 1 equivalence point, we suppose the acid is monoprotic
Step 2: Calculate moles NaOH
Moles NaOH = molarity NaOH * volume NaOH
Moles NaOH = 0.105 M * 0.02875 L
Moles NaOH = 0.00302 moles
We need 0.00302 moles of weak acid to neutralize the NaOH
Step 3: Calculate molar mass of weak acid
Molar mass = mass / moles
Molar mass = 1.168 grams / 0.00302 moles
Molar mass = 386.8 g/mol
The molar mass of the unknown acid is 386.8 g/mol
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:
The correct answer is: X is nitrogen dioxide, and Y is a metal oxide
Explanation:
Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.
The general equation is given as:
Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.
