Flashcards To Learn Chemistry
To determine the molecular formula using an empirical formula, the ratio of the molecular mass to the empirical mass must be calculated. Dividing the molecular mass of 128 grams by the empirical mass of ~16 grams (C = 12g + H = 1g, but there are 4 hydrogens in the compound) gives you approximately 8. This indicates that the empirical subscripts must be multiplied by 8 to result in the molecular formula of the unknown compound, C8H32.
A. High molecular dipole moments are characteristic of nonpolar molecules
B. Polar solutes tend to be more soluble in nonpolar solvents
C. The polarity of a molecule is dependent on its 3-D structure
D. All polar molecules are capable of hydrogen bonding
E. London dispersion interactions are usually stronger than dipole-dipole interactions
A molecule’s dipole is an electric dipole with an inherent electric field (not be confused with a magnetic dipole which generates a magnetic field). Molecules can have dipole moments due to non-uniform distributions of positive and negative charges on the various atoms. For diatomic molecules there is only one (single or multiple) bond so the bond dipole moment is the molecular dipole moment, with typical values in the range of 0 to 11 D. For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of all individual bond dipole moments.
A. Salt, NaCl.
B. Ethanol, CH3CH2OH.
C. Benzene, C6H6.
D. Hydrogen chloride, HCl.
E. Acetic acid, CH3COOH.
The law of solubility is that “Like dissolves Like”, that is non-polar solvents dissolve non-polar substances and polar solvents dissolve polar and most ionic substances.
Water is a polar solvent thus:
Salt, NaCl – ionic and very soluble in water, sea water.
Ethanol, CH3CH2OH – polar molecular (-OH) will dissolve in water, beer/wine/spirits.
Benzene, C6H6 – non-polar molecular will not dissolve in polar water.
Hydrogen chloride, HCl – polar molecular will dissolve in water, hydrochloric acid.
Acetic acid, CH3COOH – polar molecular (-COOH) will dissolve in water, vinegar.
The treatment of alkenes with acidic potassium permanganate under high temperatures results in oxidative cleavage. The nature of the products are dependent on alkene substitution, but all with possess a carbonyl functionality (carboxylic acid, ketone, CO2). Looking at the product, one can identify the two carbon atoms of the original alkenyl compound, and work back accordingly. Finally, standard IUPAC naming will apply.
A. Mixture A would have a higher boiling point
B. Mixture A would have a lower boiling point
C. Mixture B would have a lower boiling point
D. The mixtures have the same boiling point
E. The mixtures would vaporize at the same rate
The two mixtures have the same concentration a.k.a. molality. However, they do have differing ionic solutes. Mixture A has an ionic solute that would dissociate to form two ions (K+ and Cl-) whereas Mixture B has an ionic solute that forms three ions (Mg2+ and 2Cl-). In the equation:
?T(boiling) = m x Kb x i, the “I” or vant Hoff factor would be the only variable. The higher the vant Hoff factor, the greater the effect on the boiling point and/or freezing point. The von hoff factor for Mixture A is 2, but for Mixture B is 3. Therefore, Mixture B would have the greater affect on the boiling point, making it higher than normal. Between the two mixtures, B would have a higher boiling point than mixture A.
A. The two compounds are mirror images of one another
B. The two compounds can be separated by ordinary physical chemical separation methods
C. The two compounds have the same melting points
D. All of the above
Cis and trans-1,2-dibromocyclopropane share a diastereomeric relationship; that is, the molecules are stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers (i.e. same connectivities) of a compound have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other. Accordingly, diastereomers are distinct molecules with different chemical properties, and can be separated by ordinary physical means. Recall that enantiomers are mirror images of each other and are identical in all physical aspects except for their direction of rotation of plane polarized light.
The valence shell is the outer shell of an element. It is the electrons in the valence shell that interact to form compounds, either ionic or covalent. If the valence shell ends in the 4th energy level, this means the compound must be in the fourth row of the periodic table, therefore eliminating answer options A, B, and E. The valence shell shows a total of 7 valence electrons, 2 from the s and 5 from the p. The only element that has 7 valence electrons and is in the 4th row is Bromine (Br).
An approximated 1H NMR spectrum of 1,1-dibromopropane (CH3CH2CHBr2) is shown below:
In a reaction between Mg and Ar, which of the following is most likely to occur?
A. no reaction would occur
B. a precipitate would form
C. a gas would be produced
D. a color change would occur
E. an aqueous solution would form
Noble gases are inert and do not react under normal conditions. Therefore, the reaction would not take place. In order for a gas to be produced, a chemical reaction would need to occur. The same is true for the color change and the formation of a precipitate as these signal that a chemical change has happened. However, these would not form/occur in this case because the two reactants would not react chemically.
Gestrinone is a synthetic steroid hormone that is marketed as a treatment for endometriosis. What is the index of hydrogen deficiency (IHD)-or degree of unsaturation-for gestrinone?
Recall that the degrees of unsaturation corresponds to the sum of pi bonds and/or rings present in a molecule. Each ring and each pi bond contributes one degree of unsaturation; triple bonds, possessing one sigma and two pi bonds, contribute two degrees of unsaturation. Gestrinone has four rings (4), three alkenes (3), one carbonyl (1) and one alkyne (2). Thus IHD = 4 + 3 + 1 + 2 = 10
A. (45.1 grams)/( 0.302 moles)
B. (0.65(.302) grams)/(45.1 moles)
C. (45.1 grams)/(0.65 moles)
D. (0.65 grams)/0.302 moles)
E. (45.1 grams)/(0.65(.302) moles)
To determine the molar mass, the molarity equation will be needed.
Molarity = (moles of solute)/(Liters of solution)
Plugging in the values from the problem results in:
0.65 = (moles of solute)/(.302 L)
Note: The amount of the solution is given in terms of grams of water, this can be converted to mL by using the density of water (1.0 grams/mL) and then converting the mL into liters. Cross-multiplying gives the following:
0.65(.302) = moles of solute
The molar mass of a compound is in grams per mole. Therefore the molar mass of the unknown solute would be 45.1 grams of solute (from the question) divided by the number of moles just calculated 0.65(.302). Putting the pieces together results in a molar mass of the unknown solute of:
B. Trigonal Planar
D. Trigonal Planar-Bent
Cyanide is CN-1 and this compound has 10 valence electrons to use to bind together. The only orientation that works with these elements is with a triple bond between the C and N, with each element having a single pair of non-bonded electrons. This would give rise to a linear molecular geometry.
E. The reaction produces mixtures of all four.
The bromonium ion is formed when alkenes react with bromine. When the ? cloud of the alkene (nucleophile) approaches the bromine molecule (acting as an electrophile), the ?-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion. The species formed is a high-energy intermediate along the reaction path. It will react further with some nucleophile?methanol in this case. The electronic structure of the bromonium ion has its consequences in the stereochemistry of the reaction. Nucleophilic attack takes place from the bottom face (back-side to the “departing” bromide) on the more substituted carbon. The back-side attack leads to the anti addition (Br and Nu groups end up on the opposite faces of the starting double bond), and the nucleophile adds to the more substituted carbon (Markovnikov’s rule). Structure III corresponds to the above. Structure II may form, but only in very miniscule amounts
A. 22.4 liters.
B. 18.5 liters.
C. 11.2 liters.
D. 5.91 liters.
E. 1 liter.
This is solved using the ideal gas law. However, it should be recognized that 1 mol of any gas at 1 atm and 273K (or 0 deg Celsius) is approximately 22.4 L in volume. If this is not recognized, it can still be calculated from the ideal gas law, PV = nRT. We recognize that if pressure doubles while temperature and mols remain the same, the volume should be halved. Thus, the new volume is 11.2 L, which is a difference of 11.2 liters, or answer C.
Dynemicin A has two aromatic systems, three carbonyl substituents (two ketones, one carboxylic acid), two alkenes and two alkynes. Total = 15. Recall that alkynes possess two pi bonds in addition to one sigma bond, therefore C is not correct.
A. Change the water solubility of an unknown compound.
B. Change the boiling point of an unknown compound.
C. Change the molecular weight of an unknown compound.
D. Change the optical activity of an unknown compound.
E. Change the melting point of an unknown compound.
Adding acid or base to a mixture of unknown compounds is designed to change the water solubility. The loss or gain of protons will make the compound become ionic in nature, and thus soluble in water. On the other hand, compounds that are already ionic may become neutrally charged, resulting in lower solubility in water.
This is an example of an ionic compound. For every 1 iron atom, 2 sulfate ions are needed. If sulfate has a charge of -2 and 2 sulfates are present, then this results in a charge of -4. Since the ionic compound in the question is neutral, there should not be a charge left on the compound. Therefore, if sulfate has a -4 (is looking to gain 4 electrons), then the iron must have a +4 charge (be willing to transfer 4 electrons).
D. cyclopentanecarboxylic acid.
Beta-keto carboxylic acids undergo decarboxylation through ring formations. The carbonyl bond of the ketone will react with the alcohol group of the carboxylic acid to form a ring and release the carboxylic acid as CO2. The remaining enol will tautomerize into a carbonyl group again, leaving cyclopentanone. Notice a cyclopentaldehyde can’t even be a possible answer, aldehydes cannot exist in cyclo compounds.
C. nuclear decay.
D. nuclear fission.
E. nuclear fusion.
Nuclear decay involves the splitting of an unstable nucleus and emitting a radioactive particle. This process also decreases the energy of the original element. Nucleation though a chemistry term is related to the process of making a crystal from a supersaturated solution. A combustion reaction does generally result in the release of energy, but is not a mechanism used to make a nucleus more stable. Nuclear fusion is the process of adding elements together, increasing their energy, not decreasing their energy.
Quinine has four centers of chirality (highlighted with asterisks). Thus, in principle, quinine has 24 (2 x 2 x 2 x 2 = 16) stereoisomers.