(a)
Interpretation:
Between
Concept Introduction:
Boiling point:
Boiling is a form of evaporation where the conversion from the liquid state to vapor state occurs within the body of the liquid through bubble formation.
A normal boiling point is the temperature at which the liquid boils under a pressure of
London force:
London force is a type of weak intermolecular force that has an effect on boiling point. This weak temporary force occurs between two atoms or molecules (polar or non-polar)
The cause of London force is the temporary uneven distribution of electrons causing induced polarity. The strength of the London force depends on how easily a molecule can be distorted or polarized by the polarity present in another molecule. For the large molecules the outermost electrons are far from nucleus and as the result the attractive forces of nucleus acting on them are very weak and hence the outermost electrons can be easily polarized. Hence the induced polar molecule and the polar molecule will attract each other more.
Thus the molecules having higher mass have higher boiling point.
(b)
Interpretation:
Between
Concept Introduction:
Boiling point:
Boiling is a form of evaporation where the conversion from the liquid state to vapor state occurs within the body of the liquid through bubble formation.
A normal boiling point is the temperature at which the liquid boils under a pressure of
Hydrogen bonding:
Hydrogen bonding is an intermolecular force. This is an extra strong dipole-dipole interaction between a hydrogen atom covalently bonded to a small electronegative atom mainly
The vapor pressure of liquids having significant hydrogen bonding are much lower than those liquids having no hydrogen bonding. Because of presence of extensive hydrogen bonding the liquid molecules face difficulty to escape from the condensed state and additional energy is required to break the hydrogen bonds and go to the vapor state and as a result the liquids having extensive hydrogen bonding have higher boiling points.
Dipole-dipole interaction:
Dipole- dipole interactions are the strong intermolecular interaction that causes between two polar molecules.
In polar molecules there will be a force of attraction between the positive end and the negative end. Hence the molecules having higher electronegativity difference i.e. having more polarity will have higher dipole-dipole intermolecular attraction acting between them and as a result the boiling point will increase.
(c)
Interpretation:
Between
Concept Introduction:
Dipole-dipole interaction:
Dipole- dipole interactions are the strong intermolecular interaction that causes between two polar molecules.
In polar molecules there will be a force of attraction between the positive end and the negative end. Hence the molecules having higher electronegativity difference i.e. having more polarity will have higher dipole-dipole intermolecular attraction acting between them and as a result the boiling point will increase.
(d)
Interpretation:
Between
Concept Introduction:
Hydrogen bonding:
Hydrogen bonding is an intermolecular force. This is an extra strong dipole-dipole interaction between a hydrogen atom covalently bonded to a small electronegative atom mainly
The vapor pressure of liquids having significant hydrogen bonding are much lower than those liquids having no hydrogen bonding. Because of presence of extensive hydrogen bonding the liquid molecules face difficulty to escape from the condensed state and additional energy is required to break the hydrogen bonds and go to the vapor state and as a result the liquids having extensive hydrogen bonding have higher boiling points.
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Chapter 7 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Using line angle formulas, draw thestructures of and name four alkanes that have total of 7carbons, one of which is tertiary.Please explain this in detail and can you also explain how to approach a similar problem like this as well?arrow_forwardUsing dashed line wedge projections drawthe indicated compounds and indicate whether thecompound you have drawn is R or S.(a) The two enantiomers of 2-chlorobutane. Can you please explain your steps and how you would approach a similar problem. Thank you!arrow_forward5) There are no lone pairs shown in the structure below. Please add in all lone pairs and then give the hybridization scheme for the compound. (8) 10,11 7) 1.2.3 H 4 | 14 8) COC 12 13 H 16 15 H7 9) - 5.6 C 8 H 10) H 1). 2) 3)_ 11) 12) 13) 4)_ 14) 5) 15) 16) 6)arrow_forward
- The sum of the numbers in the name of isA. 11; B. 13; C. 10; D. 12; E. none of the other answers iscorrect. I believe the awnser should be E to this problem but the solution to this problem is D 12. I'm honestly unsure how that's the solution. If you can please explain the steps to this type of problem and how to approach a problem like this it would be greatly appreciated!arrow_forwardConsider the following data for phosphorus: g atomic mass 30.974 mol electronegativity 2.19 kJ electron affinity 72. mol kJ ionization energy 1011.8 mol kJ heat of fusion 0.64 mol You may find additional useful data in the ALEKS Data tab. Does the following reaction absorb or release energy? 2+ + (1) P (g) + e → P (g) Is it possible to calculate the amount of energy absorbed or released by reaction (1) using only the data above? If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (1): Does the following reaction absorb or release energy? 00 release absorb Can't be decided with the data given. yes no ☐ kJ/mol (²) P* (8) + + + e →>> P (g) Is it possible to calculate the amount of energy absorbed or released by reaction (2) using only the data above? If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (2): ☐ release absorb Can't be decided with the data given. yes no kJ/mol аarrow_forwardThe number of hydrogens in an alkyne that has a main chain of 14carbons to which are attached a cyclobutyl ring, a benzene ring, an–OH group, and a Br is A. 34; B. 35; C. 36; D. 24; E. 43arrow_forward
- Hello! I have a 500 Hz H-NMR for 1,5-bis-(4-methoxyphenyl)-penta-1,4-dien-3-one. I need to label the signals with the corresponding H's. Then, find out if the two alkenes are cis or trans by calculating the J values. I believe that I have the H-NMR labeled correctly, but not sure if I got the J values correct to determine if the two alkenes in the compound will make the compound cis or trans.arrow_forwardWhat is the only possible H-Sb-H bond angle in SbH3?arrow_forwardpls helparrow_forward
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