Answer:
Therefore, the boiling points of the alkanes increase with molecular size. For isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors.
The boiling points of the normal alkanes increase with increasing molecular weight (Table 3.3). As the molecular weight increases, London forces increase because more atoms are present to increase the surface area or the molecules.
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Rutherford is very well-known for his gold foil experiment. He used alpha rays to strike the gold foil. Before this he also observed radioactive decay. But in the gold foil experiment, he discovered the nucleus and the proton. So, the acceptable answers are: <span><em>emissions of radioactive elements and the proton</em>.</span>
Answer:
Correct answers: 2 and 3
Explanation:
1- correct would be: Isolation of ibuprofen is not dangerous, but it is necessary because only one enantiomer has effect on interaction with biologic <em>diana</em>
<em>2: Correct! This property of diastereomeric salts (differing solubilities) is really useful for the isolation of the original enantiomers</em>
<em>3: Correct! we can only observe their properties, like polirized light rotation or separation in an assimetric column for chromatography.</em>
4: correct would be: diastereomeric salts do not rotate light, they have lost the property of anantiomers that originated them
Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
All cells need oxygen in order to function
