Carbon dioxide and helium which has a higher van der waals force of attraction between molecules
1 answer:
Answer:
Carbon dioxide
Explanation:
Neither helium nor carbon dioxide has a molecular dipole, so their strongest van der Waals attractive forces are London forces.
Helium is a small spherical atom with only a two electrons, so its atoms have quite weak attractions to each other.
CO₂ is a large linear molecule. It has more electrons than helium, so the attractive forces are greater. Furthermore, the molecules can align themselves compactly side-by-side and maximize the attractions (see below).
For example. CO₂ becomes a solid at -78 °C, but helium must be cooled to -272 °C to make it freeze (that's just 1 °C above absolute zero).
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Answers:
1. 3-ethyl-3-methylheptane; 2. 2,2,3,3-tetramethylpentane; 3. hexa-2,4-diene.
Explanation:
<em>Structure 1 </em>
- Identify and name the longest continuous chain of carbon atoms (the main chain has 7 C; ∴ base name = heptane).
- Identify and name all the substituents [a 1C substituent (methyl) and a 2C substituent (methyl).
- Number the main chain from the end closest to a substituent.
- Identify the substituents by the number of the C atom on the main chain. Use hyphens between letters and numbers (3-methyl, 3-ethyl).
- Put the names of the substituents in alphabetical order in front of the base name with no spaces (3-ethyl-3-methylheptane)
<em>Structure 2</em>
- 5C. Base name = pentane
- Four methyl groups.
- Number from the left-hand end.
- If there is more than one substituent of the same type, identify each substituent by its locating number and use a multiplying prefix to show the number of each substituent. Use commas between numbers (2,2,3,3-tetramethyl).
- The name is 2,2,3,3-tetramethylpentane.
<em>Structure 3 </em>
- Identify and name the longest continuous chain of carbon atoms that passes through as many double bonds as possible. Drop the <em>-ne</em> ending of the alkane to get the root name <em>hexa-</em>.
- (No substituents).
- Number the main chain from the end closest to a double bond.
- If there is more than one double bond use a multiplying prefix to indicate the number of double bonds (two double bonds = diene) and use the smaller of the two numbers of the C=C atoms as the double bond locators (2,4-diene)
- Put the functional group name at the end of the root name (hexa-2,4-diene).
<em>Note</em>: The name 2,4-hexadiene is <em>acceptable</em>, but the <em>Preferred IUPAC Name</em> puts the locating numbers as close as possible in front of the groups they locate.
