The full range of essential dietary amino acids is provided by all of the following except Question 25 options: legumes beans with rice peanut butter on whole wheat bread poultry

## The full range of essential dietary amino acids is provided by all of the following except Question 25 options: legumes beans with rice peanut butter on whole wheat bread poultry

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determine the range of the distance a for which the reaction at B does not exceed 100 Ib downward or 200 IB upward.

## determine the range of the distance a for which the reaction at B does not exceed 100 Ib downward or 200 IB upward.

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Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm. A. 3 B. 4 C. 2 + D. 6 E. 7

## Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm. A. 3 B. 4 C. 2 + D. 6 E. 7

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Critical Thinking: Comprehensive Sexual Education versus Abstinence-Only The Bush Administration spent over 175 million dollars annually on abstinence-only sex education programs. These programs could only discuss the failure rates of other methods, nothing more (Ott and Santelli, 2007). Comprehensive programs educate students on a range of contraceptive options – including abstinence. Two separate studies now indicate that comprehensive programs delay the start of sexual activity and cut teen pregnancy when compared to abstinence-only education programs (Ott and Santelli, 2007; Center for the Advancement of Health, 2008).

## Critical Thinking: Comprehensive Sexual Education versus Abstinence-Only The Bush Administration spent over 175 million dollars annually on abstinence-only sex education programs. These programs could only discuss the failure rates of other methods, nothing more (Ott and Santelli, 2007). Comprehensive programs educate students on a range of contraceptive options – including abstinence. Two separate studies now indicate that comprehensive programs delay the start of sexual activity and cut teen pregnancy when compared to abstinence-only education programs (Ott and Santelli, 2007; Center for the Advancement of Health, 2008).

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Materials Chemistry for Engineers 1. In the van der Waals corrections to the Ideal Gas Law: (P + a/V2)(V – b) = nRT (a) What do a and b correct for from the Ideal Gas Law? (b) How would one determine a and b experimentally? Describe a proposed experiment and data analysis method for your experiment. 2. (a) What are the assumptions of the Ideal Gas Law? How did van der Waal modify these assumptions to come up with his equation of state? (b) what is an equation of state, in general? Describe in your own words. 3. Given the following data: Material a b_____ (l2.atm/mole2) (l/mole) N2 1.39 0.03913 NH3 4.17 0.03107 Aniline 26.50 0.1369 Benzene 18.00 0.1154 (a) Plot P vs. T for each gas using the van der Waals equation of state. Assume that you have a 1 liter volume and 1 mole of gas and plot the temperature on the x-axis from room temperature to 1400 K (pressures should range from about 0 atm to about 120 atm, depending on the gas). Plot the Ideal Gas Law with the other data on one plot. Are the interactions between molecules attractive or repulsive at low temperature? How do you know? What is happening with the gases at high temperature? Is one of the gases different from the others at 1400 K? (b) Discuss the nature of the intermolecular interaction that creates the deviation from ideality for each material. Are there induced dipole-induced dipole interactions, iondipole interactions, etc. for each of the different gases? Draw their chemical structures. 4. Ethane (CH3CH3) and fluoromethane (CH3F) have the same number of electrons and are essentially the same size. However, ethane has a boiling point of 184.5 K and fluoromethane has a boiling point of 194.7 K. Explain this 10 degree difference in boiling point in terms of the van der Waals forces present. Bonus, what is the size of each molecule? Show your calculation/sources.

## Materials Chemistry for Engineers 1. In the van der Waals corrections to the Ideal Gas Law: (P + a/V2)(V – b) = nRT (a) What do a and b correct for from the Ideal Gas Law? (b) How would one determine a and b experimentally? Describe a proposed experiment and data analysis method for your experiment. 2. (a) What are the assumptions of the Ideal Gas Law? How did van der Waal modify these assumptions to come up with his equation of state? (b) what is an equation of state, in general? Describe in your own words. 3. Given the following data: Material a b_____ (l2.atm/mole2) (l/mole) N2 1.39 0.03913 NH3 4.17 0.03107 Aniline 26.50 0.1369 Benzene 18.00 0.1154 (a) Plot P vs. T for each gas using the van der Waals equation of state. Assume that you have a 1 liter volume and 1 mole of gas and plot the temperature on the x-axis from room temperature to 1400 K (pressures should range from about 0 atm to about 120 atm, depending on the gas). Plot the Ideal Gas Law with the other data on one plot. Are the interactions between molecules attractive or repulsive at low temperature? How do you know? What is happening with the gases at high temperature? Is one of the gases different from the others at 1400 K? (b) Discuss the nature of the intermolecular interaction that creates the deviation from ideality for each material. Are there induced dipole-induced dipole interactions, iondipole interactions, etc. for each of the different gases? Draw their chemical structures. 4. Ethane (CH3CH3) and fluoromethane (CH3F) have the same number of electrons and are essentially the same size. However, ethane has a boiling point of 184.5 K and fluoromethane has a boiling point of 194.7 K. Explain this 10 degree difference in boiling point in terms of the van der Waals forces present. Bonus, what is the size of each molecule? Show your calculation/sources.

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Operational amplifiers are often used to amplify a sensor output. This problem will walk you through the design of a simple temperature measuring device based on a platinum wire sensor. Goal: Build a circuit that will provide a calibrated output between .32V and 2.12V for temperatures sensed between 0°C and 100°C. (The final circuit can be seen at the end of this homework but we will work out each stage in turn.) The platinum wire sensor has a resistance of 100Ω at 0°C and 138.5Ω at 100°C, or a change of 0.385Ω/°C. (The arrow through the resistor in the circuit indicates it is a variable resistor.) A 0.5mA source is used to excite the platinum wire resistor to obtain a voltage. The first stage of our circuit will be to buffer the output of the sensor so we do not load the sensor circuit by drawing off any of the .5mA current to the op amp. A. (10 points) What is V1 when the temperature is 0°C, 1°C, 20°C and 100°C? (Use at least four decimal points.) B. (10 points)The output voltage of the resistor changes by I*ΔRT where I =0.5mA and ΔRT = 0.385Ω/°C. It is too small, so we need amplify this so the V2 output in the second stage of this circuit will be 5mV per degree using a non-inverting amplifier. So we want 5mV = (I*ΔRT * gain) per degree centigrade. What is the required gain for this circuit? Choose values of R1 and R2 between 1k and 100kΩ to achieve this. Choose Rin to be 1K-10kΩ. C. (10 points) What is V2 for a temperature of 0°C and for 1°C. What is the difference between the two voltages? (Hint: The difference should be exactly 5mV! The resistance of the platinum wire will be 100.385Ω @ 1°C.) D. (10 extra credit points) We would like for the output voltage, V2, to be 0V when the temperature is 0°C. This can be done by adding a third stage with an offset voltage in the circuit below. Find Voffset so that VC = 0V when the temperature is 0°C. Let R3 =R4 = R5 and pick appropriate values of the resistors between 1k and 10kΩ. (Hint: V2= the voltage when the temperature is 0°C you found in part C. Find Voffset so that Vc =0V. Superposition may a good technique to use here. You can analyze the circuit when V2=0 and the offset is activated and then you can analyze the circuit when V2 = the value from part C and the offset voltage is zero.) What is Voffset? What values did you chose for the resistors? What is VC when the temperature is is 0°C, 1°C, 20°C and 100°C? E. The voltage VC should now be 0V when the temperature is 0°C and increase by 10mV for every degree centigrade. We need to multiply this by 1.8, the factor to convert a degree Centigrade to a degree Fahrenheit. The output of this stage, V4, should range from 0V to 1.80V and then we will add an offset to change the VF output range to 0.32V to 2.12V in the last stage. The following circuit can be used. Note that this circuit uses inverting amplifiers instead of non-inverting amplifiers. (10 extra credit points) Find the correct resistor values for R7, R8 so that V4 will range between 0V to -1.80V when the temperature is sensed between 0°C and 100°C. (10 extra credit points) Find Voffset2 and then determine VF at 0°C, 1°C, 20°C and 100°C so the VF will range from 0.32V to 2.12V. (Hint: When VC = 0V at 0°C, the V4 output of this stage will also be 0V. Determine the offset voltage so VF = 0.32V. Choose R9 = R10 = R11 = 1kΩ, so at 0°C, VF = 0.32V = VR10 with the current going from the output back through R10 to zero volts, then down through R11 and the Voffset2 source. Since VR10=VR11=0.32V, determine Voffset2. When the temperature is 100°C the output should be 2.12V.)

## Operational amplifiers are often used to amplify a sensor output. This problem will walk you through the design of a simple temperature measuring device based on a platinum wire sensor. Goal: Build a circuit that will provide a calibrated output between .32V and 2.12V for temperatures sensed between 0°C and 100°C. (The final circuit can be seen at the end of this homework but we will work out each stage in turn.) The platinum wire sensor has a resistance of 100Ω at 0°C and 138.5Ω at 100°C, or a change of 0.385Ω/°C. (The arrow through the resistor in the circuit indicates it is a variable resistor.) A 0.5mA source is used to excite the platinum wire resistor to obtain a voltage. The first stage of our circuit will be to buffer the output of the sensor so we do not load the sensor circuit by drawing off any of the .5mA current to the op amp. A. (10 points) What is V1 when the temperature is 0°C, 1°C, 20°C and 100°C? (Use at least four decimal points.) B. (10 points)The output voltage of the resistor changes by I*ΔRT where I =0.5mA and ΔRT = 0.385Ω/°C. It is too small, so we need amplify this so the V2 output in the second stage of this circuit will be 5mV per degree using a non-inverting amplifier. So we want 5mV = (I*ΔRT * gain) per degree centigrade. What is the required gain for this circuit? Choose values of R1 and R2 between 1k and 100kΩ to achieve this. Choose Rin to be 1K-10kΩ. C. (10 points) What is V2 for a temperature of 0°C and for 1°C. What is the difference between the two voltages? (Hint: The difference should be exactly 5mV! The resistance of the platinum wire will be 100.385Ω @ 1°C.) D. (10 extra credit points) We would like for the output voltage, V2, to be 0V when the temperature is 0°C. This can be done by adding a third stage with an offset voltage in the circuit below. Find Voffset so that VC = 0V when the temperature is 0°C. Let R3 =R4 = R5 and pick appropriate values of the resistors between 1k and 10kΩ. (Hint: V2= the voltage when the temperature is 0°C you found in part C. Find Voffset so that Vc =0V. Superposition may a good technique to use here. You can analyze the circuit when V2=0 and the offset is activated and then you can analyze the circuit when V2 = the value from part C and the offset voltage is zero.) What is Voffset? What values did you chose for the resistors? What is VC when the temperature is is 0°C, 1°C, 20°C and 100°C? E. The voltage VC should now be 0V when the temperature is 0°C and increase by 10mV for every degree centigrade. We need to multiply this by 1.8, the factor to convert a degree Centigrade to a degree Fahrenheit. The output of this stage, V4, should range from 0V to 1.80V and then we will add an offset to change the VF output range to 0.32V to 2.12V in the last stage. The following circuit can be used. Note that this circuit uses inverting amplifiers instead of non-inverting amplifiers. (10 extra credit points) Find the correct resistor values for R7, R8 so that V4 will range between 0V to -1.80V when the temperature is sensed between 0°C and 100°C. (10 extra credit points) Find Voffset2 and then determine VF at 0°C, 1°C, 20°C and 100°C so the VF will range from 0.32V to 2.12V. (Hint: When VC = 0V at 0°C, the V4 output of this stage will also be 0V. Determine the offset voltage so VF = 0.32V. Choose R9 = R10 = R11 = 1kΩ, so at 0°C, VF = 0.32V = VR10 with the current going from the output back through R10 to zero volts, then down through R11 and the Voffset2 source. Since VR10=VR11=0.32V, determine Voffset2. When the temperature is 100°C the output should be 2.12V.)

A) At 0 deg C —> R =100 Ohm, I … Read More...