Answer:
I think the answer is forest
I think because the CH3COCH3(second picture) has a ketone(the double bond to oxygen C=O) in the chemical structure and I believe ketones lead to stronger IMF than ethers(C-O-C). I am not sure about this, but I hope this answer will at least help give you a hint as to what to look up to figure this out. I'm sorry I couldn't be of more help!
Answer:
The time taken for the cross to disappear decreases as we move down the table
Explanation:
We have to take note of three important things we can see in the table.
1) the volume of HCl was held constant
2) the volume of the thiosulphate solution was increased steadily down the table
3) the volume of water was decreased steadily down the table.
We know that the rate of reaction increases with increase in the the concentration of reactants. The more the volume of reactants compared to the volume of the volume of water present, the higher the concentration of reactants and the faster the rate of reaction. This implies that it takes a shorter time for the reaction to get to completion.
As we increase the volume of thiosulphate and decrease the volume of water, we are manipulating the reactant concentration. In this case, concentration is being increased. Hence the reaction proceeds faster and it takes a shorter time for the cross to disappear. Hence the time taken for the cross to disappear will decreased steadily down the table.
Water is oxidized and reduced at both electrodes
<h3>Further explanation
</h3>
Electrolysis utilizes electrical energy to carry out non-spontaneous redox reactions. The ions in the solution flowed by an electric current will move toward the electrode opposite the charge.
The negative ions from the solution will move towards the positive electrode, and release electrons around the positive electrode (oxidation occurs) and electrons flow to the negative pole
While around the negative electrode electron binding occurs and a reduction reaction occurs
the reaction at the cathode:
1. the reduced active metal is water, other than that the metal will be reduced
2. H⁺ of the acid will be reduced
For reactions in anode:
1. if the electrodes are not inert then the metal is oxidized
2. If inert then:
a. OH⁻ from the base will be oxidized
b. The halogen metal will oxidize
- Na⁺ (cathode)⇒metal group 1 and 2 ⇒ H₂O reduced
2H₂O(l)+2e⁻⇒ H₂(g)+2OH⁻(aq)
- SO₄²⁻ (anode)⇒acid ion(oxyacid)⇒H₂O oxidized
2H₂O(l)⇒4H⁺(aq)+O₂(g)+4e⁻