Answer:
In the electrolysis of dilute sulfuric acid, which electrolysis in aqueous solution to form hydrogen ions, H⁺, and sulfur IV ions SO₄²⁻ in the presence of H⁺ and OH⁻ ions from the water molecules
At the anode
The anode, positive electrode, attracts the negative OH⁻ and SO₄²⁻ ions where the OH⁻ gives up electrons to form water molecules and oxygen as follows;
4OH⁻ → 2H₂O + O₂ + 4e⁻
At the cathode
The positive H⁺ ions from the water molecules and the acid are attracted to the cathode where they combine with 2 electrons to form hydrogen gas as follows;
2e⁻ + 2H⁺ → H₂ (gas)
Explanation:
Answer:
0.99%
Explanation:
The Henderson-Hasselbalch equation relates the ratio of the ionized to the non-ionized form of an acid as follows:
pH = pKa + log([A⁻]/[HA])
Inserting the values for pH and pKa and solving gives:
7.4 = 5.2 + log([A⁻]/[HA])
2 = log([A⁻]/[HA])
[A⁻]/[HA] = 100/1
The percentage of the weak acid in non-ionized form (HA) is then calculated:
[HA]/([A⁻] + [HA]) = 1/(100+1) x 100% = 0.99&
<u>Answer:</u> The mass of oxygen reacted is
<u>Explanation:</u>
We are given:
Moles of propane = 9.98 mol
For the given chemical equation:
By Stoichiometry of the reaction:
1 mole of propane reacts with 5 moles of oxygen.
So, 9.98 moles of propane will react with = of oxygen.
To calculate the mass of carbon dioxide, we use the equation:
Moles of oxygen = 49.9 moles
Molar mass of oxygen gas = 32 g/mol
Putting values in above equation:
Hence, the mass of oxygen reacted is
KE = mv²/2
m=2*KE/v²
v=50 m/s
KE=500J
m=2*KE/v² =2*500/50²=1000/2500= 0.4 kg