(a)
Interpretation:
Mixture of three molecules are separated on a
Concept Introduction:
London dispersion force also called an induced dipole-induced dipole attraction is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.
Dipole-dipole interaction results when two dipolar molecules interact with each other through space, there occurs a partially negative charge of one of the polar molecules is attracted to the partially positive charge of the second polar molecule.
Ion-dipole interaction results of an electrostatic interaction between a charged ion and a molecule that has a dipole.
A hydrogen bonding is partially an electrostatic attraction between
(b).
Interpretation:
It should be determine that which molecules is most attracted to the stationary phase, and what are the forces that attract the molecule to the non-polar phase.
Concept Introduction:
London dispersion force also called an induced dipole-induced dipole attraction is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.
Dipole-dipole interaction results when two dipolar molecules interact with each other through space, there occurs a partially negative charge of one of the polar molecules is attracted to the partially positive charge of the second polar molecule.
Ion-dipole interaction results of an electrostatic interaction between a charged ion and a molecule that has a dipole.
A hydrogen bonding is partially an electrostatic attraction between
(c).
Interpretation:
It should be determine that which molecules is most attracted to the stationary phase, and what are the forces that attract the molecule to the non-polar phase.
Concept Introduction:
London dispersion: This force also called an induced dipole-induced dipole attraction is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.
Dipole-dipole interaction: Dipole interaction results when two dipolar molecules interact with each other through space, there occurs a partially negative charge of one of the polar molecules is attracted to the partially positive charge of the second polar molecule.
Ion-dipole interaction results of an electrostatic interaction between a charged ion and a molecule that has a dipole.
A hydrogen bonding is partially an electrostatic attraction between
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Chapter 11 Solutions
Chemistry & Chemical Reactivity
- A student rushed through this experiment. Describe the effect that the following procedural changes would have on the percent recovery of acetanilide. Briefly explain the basis of each answer.(a) Rather than adding 0.5-mL portions of boiling solvent to the acetanilide, the student added 5-mL portions of boiling solvent.(b) The student did not pre-heat the gravity filtration apparatus.(c) The student forgot to cool 5 mL of solvent in Part 5 and washed the crystals with room-temperature solvent.arrow_forwardWhat is the solvent being used? (Looking at the y-axis)?arrow_forwardImagine that the blue component of a given mixture moved 2.5 cm from the oigin or starting line, and the solvent had traveled 10.0cm, What is the Rf for the blue dye A .4.4 B 0.25 C. 7.5 D. 25arrow_forward
- 3. Suppose you have a mixture of saturated aqueous sodium chloride (brine) and your 2-chloro-2-methylbutane product in a separatory funnel. Use densities to predict which phase will be the top layer in the funnel. a. 2-chloro-2-methylbutane (organic phase) b. saturated sodium chloride (aqueous phase) c. hard to predict since densities are both very close to 1.18 g/mL d. there would only be one phase since the substances are misciblearrow_forwardThe melting point (mp) range of reference compound A is 320-340°C. The mp range of a 1:1 mixture of a sample of A with a sample of unknown compound B is found to be 140-300°C. Do the samples consist of identical compounds? A)Yes, because the mp of the mixture is lower than the mp of pure A. B) Not enough information is given to answer the question C) Yes, because they melt over a broad range. D) No, because the mp of the mixture is lower than the melting point of pure A.arrow_forwardEach of the student statements below is wrong. Explain why they are not correct. (a) All my compound dissolved right away in the small amount of solvent I added at room temperature, and I didn’t need to heat it at all. This means I’m going to get lots of pure compound out. (b) I did a recrystallization of naphthalene and my percent recovery was very high (99%), so I must have pure product. (c) When you’ve dissolved all your compound in hot solvent, and you’re in a hurry, it’s ok to just place it straight into the ice bath.arrow_forward
- 1. The mixture below is seperated using HPLC with a octadecylsiloxane (nonpolar) stationary phase and a methonal (CH3OH)/water mobile phase. In what ofder will the compnents come off the column. What would happen to the order in which the components come off a polar colum if pentane is used as the mobile phase? Heptane, Heptanol, Butyl propyl etherarrow_forwardWhich combination cannot be used as mobile phase in RPLC? A. Acetonitrile-cyclopentane B. Water-acetonitrile C. Acetonitrile-THF D. Water-THFarrow_forwardWhat is the approximate amount (mL) of 95% ethyl alcohol that will be required to dissolve the sulfanilamide in boiling ethanol? Note that Solubility of sulfanilamide in ethanol (mg/mL) is as follows: 14 mg at 0 °C; 24 mg at 20 °C; 46 mg at 40 °C; 88 mg at 60 °C; 210 mg at 80 °C. Remember that the sulfanilamide used in the experiment contains 5% wt of impurity (fluorenone); therefore, (357mg) X (0.05) = 17.85 mg impurities; hence = 357 mg -17.85 mg = X mg pure Sulfanilamide. So, (X mg) (1mL/210 mg) = Y mL (80 oC) minimum warm ethanol needed.arrow_forward
- 9. Any organic solvent waste generated in this experiment should be collected in your fume hood and poured into the waste container as directed. This waste should not be poured down the drain. True Falsearrow_forward1. Sodium Borohydride + Vanillin a. Compare boiling points. What does this say about purity? Melting Point (Literature) 110C- 117C Melting Point Range (Start- Fully Melted) from Experiment (C) 113C - 117C b. Calculate how many moles of vanillian was used. Use proper sigfigs. (Steps) To a 10-mL Erlenmeyer flask equipped with a stir bar add vanillin (1.300 g) followed by EtOH (2.5 mL). If the vanillin does not dissolve, gently warm the flask with your hand. In another 10-mL Erlenmeyer flask add sodium borohydride solution [3.42 M in 1.0 M NaOH] (2.5 mL). c. Calculate theoretical yeild of vanillyl alcohol in grams. Use sigfigs. (0.431g / 1.387g) x 100 = Percent Yield (%) 31.07%arrow_forward1.Each of the student statements below is wrong. You will need to explain why they are not correct. (a) All my compound dissolved right away in the small amount of solvent I added at room temperature, and I didn’t need to heat it at all. This means I’m going to get lots of pure compound out. (b) I did a recrystallization of naphthalene and my percent recovery was very high (99%), so I must have pure product. (c) When you’ve dissolved all your compound in hot solvent, and you’re in a hurry, it’s ok to just place it straight into the ice bath.arrow_forward
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