Answer:
1) Ethanol
Explanation:
If we will have <u>interactions</u> we will need more <u>energy</u> to break them in order to go from liquid to gas. If we need more <u>energy</u>, therefore, the <u>temperature will be higher</u>.
In this case, we can discard the <u>propanone</u> because this molecule don't have the ability to form <u>hydrogen bonds</u>. (Let's remember that to have hydrogen bonds we need to have a hydrogen bond to a <u>heteroatom</u>, O, N, P or S).
Then we have to analyze the hydrogen bonds formed in the other molecules. For ethanol, we will have only <u>1 hydrogen bond</u>. For water and ethanoic acid, we will have <u>2 hydrogen bonds</u>, therefore, we can discard the ethanol.
For ethanoic acid, we have 2 <u>intramolecular hydrogen bonds</u>. For water we have 2 <u>intermolecular hydrogen bonds</u>, therefore, the strongest interaction will be in the <u>ethanoic acid</u>.
The<u> closer boiling point</u> to the 75ºC is the <u>ethanol</u> (boiling point of 78.8 ºC) therefore these molecules would have <u>enough energy</u> to <u>break</u> the hydrogen bonds and to past from<u> liquid to gas</u>.
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
This would be 8.010 * 10^-2