General Chemistry Flashcards
Which compound has the bond with the smallest dipole moment?
A) H2O
B) H2S
C) H2Se
D) H2Te
D) H2Te
Dipole moment: u=qr
q=magnitude of partial charge separated across a distance r
Te is the largest, all group 16 elements, which makes H-Te the longest bond in the given compounds. However, the electronegativity difference is the smallest in the H-Te bond because electronegativity tends to decrease moving down a group (column) on the periodic table
Therefore, the H-Te bond is the longest but has the smallest dipole moment of the given compounds because the atoms in the H-Te bond have almost no difference in electronegativity, ie negligible partial charge separated across the bond
The slope of the trendline for the data shown in Figure 1 is -4911 and its y-intercept is 5.4. Which of the following is an expression for deltaG* at temperature T stated in terms of the data variables?
A) 4911R - T(5.4R)
B) 5.4R + T(4911R)
C) 4911/R -T5.4/R)
D) (5.4/R) + T(4911/R)
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A) 4911R - T(5.4R)
Given van’t Hoff equation: lnKa = (-deltaH* / R)(1/T) + (deltaS*/R)
equate to y=mx+b
-4911=(-deltaH* / R)
deltaH* = 4911R
5.4 =(deltaS/R)
deltaS = 5.4R
Would changes in the can’t Hoff plot be observed if the reaction rate were increased by adding a catalyst during the experiment?
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No, because deltaG. deltaH, and deltaS* are independent of the reaction rate
What volume of the solutions tested would contain 0.080g of ocytocin acetate?
A) 0.020 mL
B) 0.32 mL
C) 20 mL
D) 320 mL
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D) 320 mL
0.080 g oxytocin x (1000 mg/1g) x (1mL / 0.25 mg) = 320 mL
(1mL / 0.25 mg) = sample oxytocin concentration given in passage
Suppose the 10 mM MgCl2 solution used in experiment 2 is replaced with a 5 mM NaCl solution. Compared to the Cl- concentration in the 10 mM MgCl2 solution, the concentraiton of Cl- in the 5mM NaCl solution would be:
A) greater
B) the same
C) 2 times less
D) 4 times less
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D) 4 times less
NaCL solution:
(5 mmol NaCl / 1 mL solution) x (1 mmol Cl-/1mmol NaCL) = 5 mM Cl-
MgCl2 solution:
(10 mmol MgCl2 / 1 mL solution) x (2 mmol Cl- / 1 mmol MgCl2) = 20 mM Cl
Which of the following experimental changes to the initially prepared oxytocin acetate solutions will decrease the percennt of oxytocin acetate recovered in Experiment 1 after the 4-week storage period?
A) Adding more sodium citrate after the NaOH has been added
B) decreasing the initial concentration of oxytocin acetate
C) storing the solutions at 0C rather than 5C
D) increasing the citrate buffer concentration
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A) Adding more sodium citrate after the NaOH has been added
Dissociation of a weak acid: Ka = ([HA] [A-]) / [HA]
Henderson Hasselbach: pH = pKa + log ([A-] / [HA])
Adding more acid will decrease the pH, adding more base will increase the pH
In Experiment 1, the citrate buffer is made of citric acid [HA] and sodium citrate [A-]. Adding more sodium citrate to the initially prepared solutions will increase the pH of the solution.
The passage states that oxytocin is more stable at a pH of 4.5, and Figure 1 shows that at the higher pH of 6.8 (water), less oxytocin was recovered. Therefore, adding more sodium citrate will decrease the percent recovery of oxytocin acetate.
If 16 mL of the initial oxytocin acetate solution containing the 50 mM Zn2+ additive was analyzed, how many moles of oxytocin acetate (MW = 1067 g/mol) were recovered from the initial sample after 4 weeks at 50C?
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16 mL x (0.25mg / 1 mL) x (1 g / 1000 mg) x (1 mol/1067 g) x (80 mol recovered / 100 mol original ) = 3.0 x 10^-6 mol oxytocin acetate
Suppose that citric acid (H3C6H5O7) is titrated with 0.1M NaOH to form a citrate buffer solution with a pH of 4.5. What is the pH at the first equivalence point? Note: pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40.
A) Less than 3.13
B) Between 3.13 and 4.76
C) Equal to 3.13
D) Greater than 4.76
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B) Between 3.13 and 4.76
The pH at the first equivalence point must be between the pKa of the first acidic hydrogen (pKa1) and the second acidic hydrogen (pKa2)
A radioactive atom decays by 5 alpha, 3 beta minus, and 2 gamma emissions to yield 211PO. What was the original nucleus?
A) 231Pa
B) 223At
C) 231U
D) 191IR
A) 231Pa
Alpha emission
release of a helium nucleus (two protons and two neutrons). Each alpha emission reduces the atomic mass and atomic number of the atom by 4 and 2, respectively
Beta minus emission
release of an electron from the nucleus and turns a neutron into a proton; it increases the atomic number by 1. The mass of an electron is negligible compared to the nucleus, so beta emission does not alter the atomic mass
Gamma emission
the release of a high energy photon that occurs when the protons and neutrons in a nucleus change configuration. it does not alter the atomic mass or atomic number of an atom
Consider the dimerization of nitrogen dioxide at room temp:
2No2(g) –> N2O4 (g) ; Kp = 8atm^-1
If the equilibrium partial pressure of nitrogen dioxide in a container is 0.5atm, what is the partial pressure of dinitrogen tetroxide?
2 atm
Kp = (Pproducts)^m /(Preactants)^n
=PN2O4 / PNO2 x PNo2
8atm^-1 = PN2O4 / (0.5 atm)^2
PN2O4 - 2 arm
Researchers wished to mimic the conditions of the medial Golgi (ph=6.3). Which of the buffers shown in Table 1 would be best suited for this experiment?
Acetic acid : Ka = 1.7 x 10^-5
MES: Ka = 7.1 x 10^-7
HEPES : 2.8 x 10^-8
Tris: 6.3 x 10^-9
MES: Ka = 7.1 x 10^-7
this equals pka of 7.1, within +/- region of 6.3, given pH
The chemical structures of sulfur compounds H2S, SF6, S2Cl2, and S4N4 each contain only single bonds. Based on trends in the atomic radii of the elements, the compound that contains the largest bond between sulfur and an atom of another element is:
S2Cl2
The length of a sigma bond can be estimated as the sum of the atomic radii of the bonded atoms.
Because Cl has a larger atomic radius than H, N, or F, the S-Cl bond will be the longest bond with sulfur among the given compounds.
The anion SO3 2- and sulfur trioxide, SO3, have the same elemental composition. These two chemical species are:
A) the same compound, because both have the same elemental composition
B) the same compound, because both have the same number of electrons and bonding configuration
C) different compounds, because the atoms in SO3 2- are held together by ionic bonds
D) Different compounds, because SO3 2- has two more electrons and a different bonding configuration
D) Different compounds, because SO3 2- has two more electrons and a different bonding configuration
positron emission
B+ decay
Suppose that the (p-n) nuclear reaction described in the passage was attempted but resulted in the 18 O nucleus absorbing a proton within ejecting a neutron. Which of the following statements correctly describes the resulting atom that would be formed?
A) the process would form an isotope of flourine with very different chemical properties than 18F
B) the process would form an isotope of oxygen with nearly identical chemical properties
C) the process would form an isotope of flourine with nearly identical chemical properties
D) the process would form an element other than O and F with very different chemical properties
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C) the process would form an isotope of flourine with nearly identical chemical properties
If the 18 O nucleus were to absorb a proton, the atomic number would increase by 1 to a total of 9 protons. Because all atoms with 9 protons are flourine atoms, the resulting nucleus would be a flourine isotope. If no neutron was ejected following the absorption of the proton, the mass number would also increase by 1 to give a flourine isotope with a mass number of 19. The flourine cation initially formed would quickly acquire the missing valence electron from the surrounding environment.
Because 18F and 19F are isotopes of the same element with the same electron configuration, the chemical properties of these nuclei would be nearly identical.
If the initial activity from a 1-mL dose of 18F-FDG is 200 millicuries, what will be the approximate activity of the dose after 1.5 half-lives of 18F have passed?
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70 millicuries
Measured activity = (fraction remaining) x (initial activity)
fraction remaining = 35%, on graph half life = 110 minutes, therefore 1.5 half-lives = 165 minutes. At one 165 minutes, only 35% remains
Measured activity = .35 x 200 = 70
Compared to the effective nuclear charge of 18O, the effective nuclear charge of 18F is:
A) higher, because 18F and 18O have the same number of core electrons but 18F has fewer neutrons
B) Lower, because 18F has more valence electrons than 18O
C) higher, because 18F and 18O have the same number of core electrons but has 18 has more protons
D) lower, because 18F has an equal number of protons and electrons
C) higher, because 18F and 18O have the same number of core electrons but has 18 has more protons
90Sr is converted to 90Y by which of the following processes?
A) Electron capture
B) positron emission
C) gamma ray emission
D) beta minus decay
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D) beta minus decay
Electron capture
a special form of beta decay in which a nucleus absorbs an electron from an external source without emitting a particle. The absorbed electron converts a proton into a neutron, deceasing the atomic number by one
In its nuclear decay chain, 90St decays into 90Y and then to 90Zr. Which of the following elements exhibits chemical properties most similar to zirconium-90?
A) 90Nb
B) 184Hf
C) 46Sc
D) 48Car
B) 184Hf
Zirconium is in the same column as hafnium HF, and therefore any isotope of Hf will have similar chemical properties to any isotope of Zr
Physical properties (color, density, atomic radius) are not easily determined by column grouping
Fission of 235U produces multuple elements. Based on figure 1, which of the following is the atomic number of the most common fission product?
A) 38
B) 55
C) 92
D) 133
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B) 55
The tallest peak indicates the greatest fission yield , 133
Of the answer choices given, only cesium CS with an atomic number of 55 has an atomic mass near 133
