hey guys I was having problems with some of these explanations. For anyone whos taken the test or seen these problems before help me out if you can! Thanks! I was having problems getting the diagrams from the passage onto here. So if anyone has the test or has does it recently I hope you can still help
32.The wavelength range of visible light is from 400 nm to 750 nm. According to Wien's displacement law, in what approximate temperature range would a heated compound emit visible light, if you assume that its relative intensity curve is like the curve for a blackbody?
A. 1000 K to 2000 K
B. 2000 K to 4000 K
C. 4000 K to 7000 K C is the best answer. Wien's law must be manipulated so that we can solve for T in kelvins using known λmaximums, assuming (as we are instructed to do in the question) that the curve for the heated compound is like a blackbody curve. This is accomplished by dividing Wien's constant, 2.898 × 10-3 m•K, by the wavelengths in meters for red light at one end of the visible-light range and for violet light at the other. The temperature range is therefore:
or about 4000 K to 7000 K, choice C. You could also solved this using Figure 2. At 2000 K, the wavelength of maximum emission is around 1.5 × 106 m. To get up to 7.5 × 107 m, half of 1.5 × 106 m, we need to double the energy (given that cutting the wavelength in half equates to doubling the energy), so the temperature must double. This means that the visible range starts around 4000 K, according to Figure 2. The visible range is from 400 nm to 750 nm, so the temperature range should be about 4000 K to 7500 K, to be proportional. Reading the graph is faster than doing the math. The best answer is C.
39. Starting at t = 0, the size of the angle subtended by the pendulum when it is displaced initially from its resting position, as it is in Figure 1, gradually:
A. decreases from an arc of 90˚. A is the best answer. The full range of the pendulum's swing will erode gradually over time, as energy is lost from the system to friction. This diminution of a pendulum's swing is known as dampened harmonic oscillation, which means that whatever the angle may be, it will decrease over time. Choices B and D are thus eliminated. The ball traverses a full arc of 90˚ to the right on its first swing (the first half of its first cycle), because it was initially raised and displaced to the left of the equilibrium position by 45˚. The term subtended refers to the angle spanned during a complete swing of the pendulum in one direction, which in this case is 90˚. The best answer is A.
B. increases from an arc of 90˚.
C. decreases from an arc of 45˚.
D. increases from an arc of 45˚.
139.What is the isoelectric point (pI) of ADH?
A. 9.00
B. 10.25
C. 11.25 C is the best answer. Vasopressin has 3 dissociable hydrogens. One is on the N-terminus (pKa ≈ 8.0), one is on the phenolic hydroxyl (pKa ≈ 10.0), and the last is on the guanidino group of arginine (pKa ≈ 12.5). The isoelectric point (pI) is that pH at which the molecule (vasopressin) has no net charge and is in the zwitterion form. As shown by the structure of vasopressin in Figure 1, there is an overall +2 charge. What pKa must be exceeded before we get to a vasopressin molecule with a +1 overall charge? The first hydrogen to come off, if we titrate the molecule with a base is the one with the lowest pKa. The hydrogen that comes off first is the N-terminus hydrogen. Once this happens, there is no longer a positive charge on that nitrogen. The vasopressin molecule now has an overall charge of +1. The next hydrogen to come off belongs to the phenolic hydroxyl. Removal of that hydrogen places a -1 charge on the phenolic oxygen atom. At this point, vasopressin now has a -1 charge on the phenolic oxygen and a +1 charge on the guanidino group. The overall charge is zero, and we have found the zwitterion. The pI is simply the average of the two pKa values on either side of the zwitterionic form.
The best answer is C.
D. 15.25
32.The wavelength range of visible light is from 400 nm to 750 nm. According to Wien's displacement law, in what approximate temperature range would a heated compound emit visible light, if you assume that its relative intensity curve is like the curve for a blackbody?
A. 1000 K to 2000 K
B. 2000 K to 4000 K
C. 4000 K to 7000 K C is the best answer. Wien's law must be manipulated so that we can solve for T in kelvins using known λmaximums, assuming (as we are instructed to do in the question) that the curve for the heated compound is like a blackbody curve. This is accomplished by dividing Wien's constant, 2.898 × 10-3 m•K, by the wavelengths in meters for red light at one end of the visible-light range and for violet light at the other. The temperature range is therefore:
or about 4000 K to 7000 K, choice C. You could also solved this using Figure 2. At 2000 K, the wavelength of maximum emission is around 1.5 × 106 m. To get up to 7.5 × 107 m, half of 1.5 × 106 m, we need to double the energy (given that cutting the wavelength in half equates to doubling the energy), so the temperature must double. This means that the visible range starts around 4000 K, according to Figure 2. The visible range is from 400 nm to 750 nm, so the temperature range should be about 4000 K to 7500 K, to be proportional. Reading the graph is faster than doing the math. The best answer is C.
39. Starting at t = 0, the size of the angle subtended by the pendulum when it is displaced initially from its resting position, as it is in Figure 1, gradually:
A. decreases from an arc of 90˚. A is the best answer. The full range of the pendulum's swing will erode gradually over time, as energy is lost from the system to friction. This diminution of a pendulum's swing is known as dampened harmonic oscillation, which means that whatever the angle may be, it will decrease over time. Choices B and D are thus eliminated. The ball traverses a full arc of 90˚ to the right on its first swing (the first half of its first cycle), because it was initially raised and displaced to the left of the equilibrium position by 45˚. The term subtended refers to the angle spanned during a complete swing of the pendulum in one direction, which in this case is 90˚. The best answer is A.
B. increases from an arc of 90˚.
C. decreases from an arc of 45˚.
D. increases from an arc of 45˚.
139.What is the isoelectric point (pI) of ADH?
A. 9.00
B. 10.25
C. 11.25 C is the best answer. Vasopressin has 3 dissociable hydrogens. One is on the N-terminus (pKa ≈ 8.0), one is on the phenolic hydroxyl (pKa ≈ 10.0), and the last is on the guanidino group of arginine (pKa ≈ 12.5). The isoelectric point (pI) is that pH at which the molecule (vasopressin) has no net charge and is in the zwitterion form. As shown by the structure of vasopressin in Figure 1, there is an overall +2 charge. What pKa must be exceeded before we get to a vasopressin molecule with a +1 overall charge? The first hydrogen to come off, if we titrate the molecule with a base is the one with the lowest pKa. The hydrogen that comes off first is the N-terminus hydrogen. Once this happens, there is no longer a positive charge on that nitrogen. The vasopressin molecule now has an overall charge of +1. The next hydrogen to come off belongs to the phenolic hydroxyl. Removal of that hydrogen places a -1 charge on the phenolic oxygen atom. At this point, vasopressin now has a -1 charge on the phenolic oxygen and a +1 charge on the guanidino group. The overall charge is zero, and we have found the zwitterion. The pI is simply the average of the two pKa values on either side of the zwitterionic form.
The best answer is C.
D. 15.25
