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13 Dec 2019

For any real number r, let fl(r) denote the double precision representation of r in the IEEE floating point
standard. The following should illustrate that with limited precision, the laws of arithmetic do not hold.

A) Give an example of rational numbers x; y; z such that fl(fl(x + y) + z) 6= fl(x + fl(y + z)).
B) Give an example of a rational number x > 0 such that fl(x â fl(1=x)) 6= fl(1).
C) Give an example of rational numbers x; y; z such that fl(fl(x â y) â z) 6= fl(x â fl(y â z))

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Consider a dilute solution of glucose in water at 25 oC. The solution was in a horizontal tube of length 10 cm, at was prepared such that there was a linear gradation of glucose concentration from cglucose(0, 0) = 0.1M at x = 0 cm to cglucose(0, 10cm) = 0.05M at x = 10 cm. (A) What is the force (and direction) acting at the center of the tube (x = 5 cm)? Derive a formula for cglucose(0, x) for x â [0, 10]cm. (B) We wish to solve the time time dependence of cglucose(t, x) for all times t > 0. The next steps of this question will lead you through how to solve such a problem. Consider Fickâs second law of diffusion in one dimension, âc ât = D â 2 c âx2 (2) Show that a function of the form c(t, x) = exp(Î»t) Ã (Aeikx + Beâikx) satisfies Eq. (2) when Î» is related to k and D. Determine Î» for such a solution. (C) For the problem like that considered in (A) for particles confined to a a region of length L where x â (0, L) (in (A), L = 10 cm, for example) The flux at the end of the containers is equal to zero (no particles leave or enter the container). Mathematically, this imposes a boundary condition that J(x â 0â) = J(x â L+) = 0 (x â 0â means you approach x = 0 from the left and x â L+ means you approach x = L from the right). Show that these boundary conditions imply that A = B and k â¡ kn = nÏ L for n = 0, 1, 2, ...â. That is, show that solutions look like 2AeÎ»nt cos nÏx L (where we now relate Î»n to kn). (D) Spatially normalize the solutions. Letâs denote the solutions to Eq. (2) with the above boundary conditions as cn(t, x) = AË ne Î»nt cos nÏx d for n = 0, 1, 2, ...â. We wish to verify that at t = 0, these solutions are orthonormal. What the means is we want to show: Z L 0 dxcn(0, x)cm(0, x) = 0 (3) for n 6= m and Z L 0 dxc0(0, x)c0(0, x) = L(AË 0) 2 = 1 Z L 0 dxcn6=0(0, x)cn6=0(0, x) = L 2 (AË n) 2 = 1 (4) 3 (Hint: it may be easier to verify the above integrals using the relationships that cos(Î¸) = e iÎ¸+eâiÎ¸ 2 and sin(Î¸) = e iÎ¸âeâiÎ¸ 2i ). From the above orthonormality conditions, determine what An for n = 0, 1, 2, ...â. (E) Superposition principle. Verify that c(t, x) = Pâ n=0 bncn(t, x) satisfies Eq. (2) and also the boundary condition that âc(t,x) âx xâ0â = âc(t,x) âx xâL+ = 0. The coefficients, bn are simple numbers telling you how much of cn(t, x) is part of your solution for c(t, x). (F) Imposing the initial condition at t = 0. Suppose that at t = 0, c(0, x) = F +Gx for x â (0, L) and c(0, x) = 0 for x â¤ 0 and x â¥ L, i.e., there is a linear gradation to the concentration within the sample region. By the superposition principle, we can write c(t, x) as c(t, x) = X n=0 bncn(t, x) (5) At t = 0 we must have: c(0, x) = F + Gx = X n=0 bncn(0, x) (6) Using orthonormality conditions you derived above, show that bn = Z L 0 dxcn(0, x)(F + Gx) (7) Explicitly determine formulas for b0 and bn6=0. Using these expressions, finally write out your solution to c(t, x). (G) Finally, letâs go back to (A) and determine how the glucose concentration changes with time. The diffusion coefficient for glucose in water is D = 5.7 Ã 10â10 m2 s . Numerically calculate cglucose(t, x) for x â (0, L) at t = 0.5 s, t = 1s, t = 10s, and t = 1000 s. You will need to approximate your formulas for c(t, x) by taking only a finite number of terms.

An equimolar mixture of hexane and heptane is continuously produced in an oil refinery at a rate of 390lbmol/h. The alkane mixture is to be separated via continuous distillation with a design specification of 2% impurity allowable in each stream. The feed is a 2-phase mixture (75 mol% liquid and 25 mol% vapor). Assume that the liquid is ideal and that the columnâs operating pressure is 1 atm. Assume a full condenser and a partial reboiler are used.

Do the following. You will find L08C and L08D to be helpful.

1.Use http://app.knovel.com/web/toc.v/cid:kpYHACVPEH to find the relative volatility a when the temperature is such as to give about 1 atm vapor pressure for hexane, and another value at the temperature giving about 1 atm vapor pressure for heptane.

2.Use the average of the relative volatilities found in 1 to calculate y versus x, choosing the component giving y > x. (Hint: see slide 18 in L05B.) Plot this using whatever method you like. Include the 45^o x=y line. Youâll need 4 copies of this.

3.Graphically determine the minimum number of theoretical trays N_min in the column that would meet the specifications.

4.Graphically determine the minimum reflux ratio R_min .

5.Graphically determine the minimum boilup ratio R_B,min . (This corresponds to the minimum reflux ratio determined in 4, but using the operating line for the stripping section of the column.)

6.Let us consider the economics of distillation. There are two costs â construction and operating. The operating cost is primarily for coolant for the condenser and heating fluid for the reboiler. At total reflux, we get the lowest construction cost and a low operating cost, but no product. Thus, the cost per unit of product is infinite. At minimum reflux, the operating cost is a minimum but the column is infinitely high, and so again the cost per unit of product is infinite. The economic optimum condition is somewhere in between.

Estimate the economic optimum reflux ratio using the Gilliland Correlation (Figure 9.20 on
p 306 of the text) by selecting (R - R_min)/(R+1) = 0.4. Calculate R using your R_min, construct a McCabe-Thiele diagram using this R, and graphically determine the number of theoretical trays in the column. Compare this with that predicted by the Gilliland Correlation. What is the optimal feed tray, counting from the top?

Submit all of your calculations and plots.

Decomposition of Malachite

Here is what Malachite is Made Out Of

Malachite is a famous and very popular semi-precious stone. Its banded light and dark green bands are one-of-a-kind, and give it a unique ornamental quality unlike that of any other stone. The light and dark green bands are so distinctive that malachite may be one of the most easily recognized minerals by the general public. Malachite is popular in jewelry, especially Native American southwestern jewelry. The stones inlayed in silver make a nice alternative to traditional turquoise jewelry.

Malachite is a form of copper carbonate with the molecular formula Cu₂CO₃(OH)₂. When heated to around 200 ï¿½C, malachite decomposes into copper (II) oxide (CuO) and other products.

Reaction Stoichiometry

The Law of Conservation of Mass states that matter can neither be created nor destroyed. Therefore, the number of atoms of each element on the reactants side must be same as the number of atoms on the products side. This quantitative study of chemical reactions is called stoichiometry. The substances in a balanced reaction are related to each other by stoichiometric molar ratios. For example, magnesium reacts with hydrochloric acid in a 1:2 stoichiometric ratio in the balanced equation below.

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

The molar ratios can be used as conversion factors to find the amount (in moles) of one substance given the amount of another. For example, given the number of moles of HCl, the amount of MgCl₂ can be determined as shown below.

nMgCl2=nHCl x (1mol MgCl2) / (2mol HCL)

The molar ratios are used to determine the amount in moles. However, the amount of reactant or product is often measured in grams. Mass can be converted to moles and vice versa as shown below.

n = m / (MM)

m = n x MM

where n is the number of moles, m is the mass of the product in g, and MM is its molar mass in g/mol.

To observe the Law of Conservation of Mass, you can perform different types of chemical reactions. For example, you can use a decomposition reaction to observe one compound being broken down into two or more elements or new compounds. For this reaction to take place, you must break existing bonds within the starting compound. Different sources of energy such as heat, light, or electricity may be used. Through a decomposition reaction, you can observe that the atoms in the original compound are not destroyed but are rearranged to form simpler substances, such as elements or compounds.

About This Lab

In this lab, you will test if the relationship between malachite and copper (II) oxide is stoichiometric. To do this, you will heat a known mass of malachite to produce copper (II) oxide. You will then compare the amount of copper (II) oxide obtained experimentally with the amount of copper (II) oxide expected from the balanced equation for the decomposition reaction.

1.Take a crucible from the Containers shelf and place it on the workbench.

2.Take a balance from the Instruments shelf and place it on the workbench.

3.Move the crucible onto the balance. Record the mass of the empty crucible to the nearest 0.001g in your Lab Notes.

4.Move the crucible onto the workbench.

5.Take copper (II) carbonate hydroxide from the Materials shelf and add 10 g to the crucible.

6.Move the crucible onto the balance and record the mass of the crucible containing copper (II) carbonate hydroxide in your Lab Notes.

7.Take a Bunsen burner from the Instruments shelf and place it on the workbench.

8.Move the crucible onto the Bunsen burner.

9.Double-click on the crucible to open its properties. Select Show Contents and click OK. This gives you a cut-away view of the crucible's contents.

10.Turn the Bunsen burner on to low flame by clicking the knob at the bottom once. Clicking multiple times will increase the intensity of the flame until it is turned off.

11.Watch the crucible until the malachite fully decomposes. You will see a distinct color change.

12.Move the crucible onto the workbench to cool.

13.Turn off your Bunsen burner.

14.Weigh the crucible with its contents and record the mass to the nearest 0.001 g in your Lab Notes.

Clear your station.

Lab Notes:

Numbers below refer to numbers in the above âAbout this Labâ

3. Empty crucible = 88 g

5. Added 10 g of Copper (II) carbonate hydroxide to crucible

6. Crucible with Copper (II) carbonate hydroxide = 98 g

11. to Black

14. after heating crucible with its contents = 95.195

Questions I have answered:

How many grams of malachite did you add to the crucible? 10g

How many grams of product were in the crucible after heating? 7.195

By how much did the mass of the crucible's contents change after heating? -2.805

The molar mass of malachite is 221.1 g/mol. How many moles of malachite were present before the reaction? 0.04523 mol

The product remaining in the crucible after heating is copper (II) oxide. The molar mass of copper (II) oxide is 79.55 g/mol. How many moles of copper (II) oxide were produced during the reaction?

0.0945 mol

Please answer the following:

Q1 a. Which of the following represents the balanced chemical equation for this reaction?

Cu2CO3(OH)2=2CuO+H2O

Cu2CO3(OH)2=CuO+CO2

Cu2CO3(OH)2=2CuO+H2O+CO2

Cu2CO3(OH)2=CuO+H2O+CO2

b. Based on the number of moles of malachite that you started with, how many grams of water were produced? The molar mass of water is 18.0153 g/mol. Choose the closest answer.

1.815 g 0.8148 g 1.630 g 1.982 g

c. Based on the number of moles of malachite that you started with, how many moles of CO₂were produced? Choose the closest answer.

1.000 mol 0.04523 mol 0.9000 mol 0.09045 mol

d. Based on the number of moles of malachite that you started with, how many grams of CO₂were produced? The molar mass of CO₂ is 44.01 g/mol. Choose the closest answer.

1.991 g 0.4401 g 0.04523 g 3.981 g

Q2a. Based on the number of grams of CO₂ and H₂O produced during the reaction, does this compensate for the observed loss of mass?

A. No, because this violates the law of conservation of mass

B. yes, this compensates for the mass lost in the reaction

C. No, because more copper oxide was produced

D. No because mass is conserved

b. You observed that the products in the crucible weigh less than the reactants that you added. How is this possible?

A. Mass was lost due to experimental error

B. The Malachite absorbed CO2 from the air

C. CO2 and H2O were released as gases

D. Mass was converted to energy

c. What kind of a reaction took place upon heating?

A. decomposition

B. single displacement

C. synthesis

D. Double displacement

d. Baking soda is used as a leavening agent in baking. When heated, baking soda undergoes a decomposition reaction to form carbon dioxide. Given the data below, which of the following reactions occurs during baking? The molar mass of baking soda is 84.007 g/mol and the molar mass of carbon dioxide is 44.01 g/mol.

Mass of empty crucible (g) = 55.000

mass of crucible with baking soda (g) = 57.000

mass of crucible and contents after the reaction (g) =56.260

A. 2NaHCO3= 2Na+2CO2 +H2O

B. 2NaHCO3 = Na2O + 2CO2 +H2O

C. NaHCO3 = NaOH +CO2

D. 2NaHCO3 = Na2CO3 + CO2 +H2O

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