Wednesday

July 29, 2015

July 29, 2015

Posted by **3.091** on Wednesday, July 30, 2014 at 9:57am.

The decomposition of NO2 at room temperature exhibits the following variation in concentration with time:

The concentration of NO2 is expressed in mol/liter, while time is expressed in seconds.

[NO2] ln[NO2] 1/[NO2] Time (s)

0.0831 -2.4877 12.03 0

0.0666 -2.7091 15.02 4.2

0.0567 -2.8700 17.64 7.9

0.0487 -3.0221 20.53 11.4

0.0441 -3.1213 22.68 15.0

a. Express the rate of decomposition of NO2 as a function of the concentration of NO2 and determine the order of reaction. (numerical response only: 1 = 1st order, 1.5 = three halves order, 2 = 2nd order, etc.)

2 - correct

b. Determine the value of the rate constant.

0.71 - correct

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WOULD YOU WANT TO BE A SILICON RIBBON TWIRLER? (10/10 points)

A silicon ribbon measuring 100 micron thick, 5 mm wide, and 1 m long has 5 volts applied across its long dimension. The resistivity of silicon is 640 ohm-m. How much current will flow through the ribbon? Express your answer in Amps.

3.906*10^-9 - correct

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FUN WITH GALLIUM NITRIDE (10/10 points)

Determine if light with wavelength 3.87 x 10-7 m incident on gallium nitride (GaN) can generate electrons in the conduction band. Gallium nitride has a band gap of is 3.2 eV.

yes - correct

Photons with wavelengths below what value will generate electrons in the conduction band? (Express your answer in meters.)

3.872*10^-7 - correct

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IT'S PLANE SIMPLE! (10/10 points)

Calculate the planar packing density (fractional area occupied by atoms) on the (110) plane of nickel at 300K.

0.5549 - correct

Calculate the linear packing density (atoms/cm) for the [100] direction in nickel at 300K.

2.837*10^7 - correct

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IDENITIFYING ELEMENTS, LIKE A TRANSLATOR (10/10 points)

Determine the chemical element that will generate [mathjaxinline]K_{\alpha} \text{ of 7.725 x $10^8$ $\dfrac{J}{mol photons}$}[/mathjaxinline]. Enter the symbol representing the element below (i.e. enter H for hydrogen, Xe for Xenon, etc.).

Cu - correct

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X-RAY DIFFRACTION OF AN UNKNOWN METAL, MIGHT BE DANGEROUS (10/10 points)

Last week you and a friend started an experiment to obtain the X-ray diffraction peaks of an unknown metal. Through these diffraction peaks you wanted to determine:

(a) whether the cell is SC, BCC, or FCC

(b) the (hkl) value of the peaks

(c) the lattice parameter a of the metal

Unfortunately, however, your friend (since he's not in 3.091) left MIT yesterday, not to return until next semester. All of the data that you could recover from the rubble in his room was the following:

sin^2(\theta) 0.120 0.239 0.480 0.600 0.721 0.841 0.956

You also know that the metal is in the cubic crystal system and the wavelength of the X-rays used is

lambda_CuK_alpha. Using the following information, determine the information you were originally interested in (a, b, and c above).

a. Is the cell SC, BCC, or FCC?

SC- correct

b. Enter the hkl value of the peaks as a list separated by commas. Do not put spaces between the values. For instance:

(100),(111),...

(100),(110),(111),(200),(102),(112),(202) - correct

c. Enter the lattice parameter a in Angstroms.

2.24 - correct

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ELASTICITY, LIKE YOUR SOCKS (10/10 points)

The crystal structure of graphite is shown below. Use the figure to answer the following questions.

Compare the Young's moduli for the directions indicated in the figure. Fill in the blank.

The modulus along a is _________ that along b

equal to - correct

The modulus along a is _________ that along c

greater than - correct

The modulus along b is _________ that along c

greater than - correct

Which of the following reasons best explain your reasoning when comparing the moduli of b and c?

The Poisson ratio is greater than 0.3 in graphite.

The two directions are crystallographically identical

Fundamental bending and stretching of bonds is different in the two directions

Van der Waals bonding has a lower bond strength than colavent bonding - correct

Stretching of covalent bonds requires greater force than bending such bonds

The opportunity to stetch rings in one direction gives one direction a lower modulus

Which of the following reasons best explain your reasoning when comparing the moduli of a and b?

Stretching of covalent bonds requires greater force than bending such bonds

The Poisson ratio is greater than 0.3 in graphite.

The two directions are crystallographically identical - correct

Fundamental bending and stretching of bonds is different in the two directions

The opportunity to stetch rings in one direction gives one direction a lower modulus

Van der Waals bonding has a lower bond strength than colavent bonding

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VACANCY IN HOTEL CALIFORNIA (10/10 points)

An activation energy of 5 eV is required to form a vacancy in a metal. At 827oC there is one vacancy for every 104 atoms. At what temperature will there be one vacancy for every 1000 atoms? Express your answer in Kelvin (K).

1150.23 - correct

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JUST SOME DIFFUSIONAL DATA CRUNCHING (10/10 points)

Consider the data below

Temperature (0C) Diffusivity (m2/s)

736 2 x 10-13

782 5 x 10-13

835 1.3 x 10-12

Given this data, determine the activation energy for the diffusion process (units of Joules per atom or molecule) in this material and the pre-exponential factor (use units of m2/s).

The activation energy is

2.928*10^-19 - correct

The pre-exponential factor is

0.000267191 - correct

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HYDROGEN STORAGE IN ALLOYS - THE FUTURE? (10/10 points)

Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be half the surface concentration after 1 hour? (Express your answer in centimeters). Assume the diffusivity of hydrogen in the alloy is 3.091x10-6 cm2/sec. Use the approximation erf(x) = x for x with values between 0 and 0.6, if appropriate. Use the Erf Table on the Course Info page for values greater than 0.6.

0.105 - correct

The activation energy for H diffusion in LaNi5 is 0.25 eV. How far will the H diffuse (where will it reach half the surface concentration) if the experiment above is repeated at 35°C? (Express your answer in centimeters).

0.124 - correct

- Chemistry -
**DrBob222**, Wednesday, July 30, 2014 at 5:02pmInteresting post but I don't see a question.