Answer:
At anode - 
At cathode - 
Explanation:
Electrolysis of NaBr:
Water will exist as:
The salt, NaBr will dissociate as:
At the anode, oxidation takes place, as shown below.
At the cathode, reduction takes place, as shown below.
The best answer for the question above would be the chloroflourocarbons or the CFCs. These chloroflourocarbons or CFCs are the ones responsible for the depletion of the ozone - which leads to leaving a hole in its layer. These gases eat out the ozone layer and allows harmful UV rays of the sun to come in the Earth.
Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
There is no reaction.
<em>Molecular equation
:</em>
K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)
<em>Ionic equation
:</em>
2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)
<em>Net ionic equation
:</em>
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>
<em>All ions cancel</em>. There is no net ionic equation.
