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>.
Answer:

Explanation:
Hello,
In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:

Best regards,
The generic solution of the reaction that occurs would be writen as water + anhydrous salt = hydrated salt
chemical equation would look like this xH2O +AB = AB.xH2O
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
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