Paper Mill Waste Proposal

My plan for disposing of the waste that comes out of the mill is actually to not dispose of it. Instead it can be recycled and treated to become a "deicer" for school sidewalks and public parks when it gets icy outside. I will desribe a quick overview of what will happen for the waste to get to this stage, then I will break it down and tell you why this works. First, the waste will be treated with hydrochloric acid. The hydrochloric acid reacts with the small amount of calcium carbonate in the waste to create a calcium chloride. Then the waste as a whole needs to be dried. Once dried, the waste, now deicer, is ready to be spread across frozen school walkways and public areas across the country to help create a safer place for our students and all citizens.

The first step in the process is to react the waste with hydrochloric acid. This is done because the waste consists of 8-12% calcium carbonate. When the hydrochloric acid reacts with the calcium carbonate it creates calcium chloride, carbon dioxide, and water. The chemical equation for this reaction is:
CaCO3(s) + 2HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
Once this reaction takes place, you will be left with a aqueous solution of calcium chloride in water. The carbon dioxide is given off into the atmosphere as a gas. To get the calcium chloride to a solid the water needs to be evaporated. This can be done by boiling off the water.

Once this is done, The waste now contains about 8-12% calcium chloride which can be used as a deicer.

To test this, I did two different experiments. The first was very basic. The plan was to fill up the tops of two ziplock container lids with water and freeze them to create a thin layer of ice similar to what would be on a sidewalk on a cold morning. Then we would take them out of the freezer, but a thin layer of the treated waste on top of one of the lids of ice, but not on the other. I will be able to observe the differences in the speed of the ice melting. This is how the test was set up:

While observing, it became obvious that the ice with the thin layer of treated waste on top of it was melting slightly faster. This cause me to believe that my theory would for sure work. I just needed another test that could statistically prove that this is correct. 

For my second test, I will do an experiment to collect information that will back up my calculations of the the freezing point depression. Freezing point depression is the difference in temperature from which a solution will freeze at compared to the temperature at which just the pure solvent will freeze at. In our case, the solvent is water, and the freezing point of water is 0°C. This means that the difference in temperature (freezing point depression) will also be the freezing point of the solution. So if the freezing point depression is below 0°C, then the treated waste will officially be successful as a deicer. To find freezing point depression, I needed a few points of data, the amount of solute (CaCl2) in moles, the molar mass of the solute, the mass of solvent, the molality of the solution, and the molal freezing point of the constant H2O (K). The equation I will follow to calculate the freezing point depression is ΔT=Km, that is, freezing point depression equals constant of solute times molality of solution. To do this equation, I needed the molality of the solution. The first step in doing this is finding the mass of the solute. This is done with the equation:

mass of solute (g) x 1 mol solute = amount of solute in moles

                             molar mass solute

I did this to my solution. The mass of the solute was 1.77g CaCl2 (I got the mass of the solute by taking the total mass of the treated waste, 17.742g and multiplying it by .10 to find 10% of the waste. I did this because the MSDS tells us that the CaCl2 takes up 8-12% of the waste, and the average of that is 10%.) and the molar mass of CaCl2 is 110.986g CaCl2. With these points of data, I was able to calculate the moles of CaCl2 in the waste using the equation above:

1.77g CaCl2 x 1 mol CaCl2 = 0.0159 moles CaCl2

                               110.985g CaCl2

I can now use the 0.0159 moles of CaCl2 to calculate the molality of the solution. The equation used to find molality is:

amount of solute (mol) x 1000g water = molality of solution

mass of solvent (g)          1 kg water

I did this for my solution. The amount of solute in moles was 0.0159 moles CaCl2 and the mass of solvent was 74.905g H20. With these points of data, I was able to calculate the molality of the solution CaCl2 and water in the waste using the equation above:

0.0159 mol CaCl2 x 1000g H20 = 0.2122 molality, which rounds to 0.2m

       74.905g H2O               1kg H2O

Now that I have the molality, I can apply it to the freezing point depression equation, ΔT=Km.

ΔT = (-1.86°C/m) x 0.2m CaCl2 = -0.372°C

-0.372°C is the freezing point depression. This is the difference in temperature from the freezing point of the solute, to the freezing point of the solution. Since the solute is H20 and the freezing point of H20 is 0°C than to find the freezing point of the solution, I had to subtract 0.372°C from 0°C to get a freezing point of -0.372°C. This means that a portion of 74.905g of water will only stay frozen until it is -0.372°C is it is mixed with 17.742g of the treated waste (which contains 1.77g CaCl2) compared to the water staying frozen until it is 0°C with just pure H2O. This theoretically proves that the treated waste will work as a deicer! Now for my experiment that backs this up. The experiment was to test the difference in freezing water with the treated waste in it, and plain water itself. I took 80mL of water in two different beakers, and added 17.742g of treated waste to it (17.742g of treated waste contains about 1.77g of CaCl2, the deicing component). These to beakers were then placed in the freezer with a temperature probe in each one, so that i was able to read the temperature of the water in the beakers from outside the freezer.

I then observed the temperature of the beakers every 5 minutes for 1 hour and 5 minutes.

                                                                           All temperature readings are in °C

Times taken:	Temp of Water	Temp of Water + CaCl2
11:55 (starting temp, before freezer)	24.0	24.9
12:00	16.2	19.4
12:05	13.5	16.5
12:10	9.4	12.8
12:15	5.9	9.5
12:20	2.8	6.6
12:25	2.1	4.0
12:30	0.7	1.2
12:35	0.3	0.5
12:40	0.3	-0.1
12:45	0.3	-0.4
12:50	0.3	-0.5
12:55	0.3	-0.6
1:00	0.3	-0.6

This data shows that when freezing the water with CaCl2 in it keeps decreasing in temperature even after the plain water has plateaued. This means that it still was not frozen and the water was still decreasing temperature as a liquid.             

After the two tests, and one theoretical calculation, I have concluded that by treating the waste with HCl you can indeed create a deicer that will be effective on school sidewalks and other public areas to keep people safe.

Citations:
http://www.ehow.com/about_6320079_pure-calcium-chloride-deicer.html
Sarquis, Mickey and Sarquis, Jerry. Modern Chemistry, Houghton Mifflin Harcourt, 2012.
All other information came from Dr. Lee

Specific Heat

Purpose: The purpose of this lab was to determine which metal we were given by determining the specific heat.

*I HAD SAMPLE #1*

Background: For this lab I will need to heat the metal I am given, place it in a cup of water, and calculate the temperature change. I will need the mass of the water, mass of the metal, the temp of the water before I put the metal in and the temp of the water after I put the metal in for 2 mins for my calculations. Once I do the calculatons, I will compare the specific heats that I calculated to the specific heat of the metals give to us to determine which metal I have. The specific heats of water, and the metals are...

• Water- 4.184J/g*°C
• Al- 0.897J/g*°C
• Brass- 0.385J/g*°C
• Cu- 0.385J/g*°C
• Lead- 0.129J/g*°C
• Stainless Steel 0.490J/g*°C
• Zn- 0.390J/g*°C

Calculations: Symbols used- q=heat, m=mass, c=specific heat, ΔT=change in temp
From my data I will have mass, specific heat, and temp change of water, so all I need is heat. Here is the calc to find the heat. It is, heat equals mass times specific heat times temp gained.
Calc for water: qw=mwcwΔTw
Now that I have heat, I can plug that in with the mass and temp change of metal to find the metals specific heat. Here is the calc to find the specific heat of my metal. It is, specific heat equals heat divided my mass times temp gained.
Calc for metal: cm=qm/mmΔTm
Now that I have the specific heat, I can compare it to the specific heats of the metals given to us and determine which metal I have.

Data: To get a more accurate result, I did the test 4 times to get 4 sets of data which I will be able to compare the calculations.


Analysis: Now with this data I will be able to preform the calculations that I explained in the background for each of my tests.

Test 1: Finding Heat: q=87.87g x 4.184J/g*°C x 5.5°C
                                  q=2022.064 Joules
            Finding Specific Heat Of Metal: c=2022.064 Joules / 33.767g x 69.1°C
                                                                c=0.867J/g*°C
                       ***Specific heat of metal in test 1 is 0.867J/g*°C

Test 2: Finding Heat: q=97.71g x 4.184J/g*°C x 5.3°C

                                  q=2166.739 Joules

            Finding Specific Heat Of Metal: c=2166.739 Joules / 33.767g x 70.6°C

                                                                c=0.909J/g*°C
                       ***Specific heat of metal in test 2 is 0.909J/g*°C

Test 3:  Finding Heat: q=87.00g x 4.184J/g*°C x 6.0°C

                                  q=2184.048 Joules

             Finding Specific Heat Of Metal: c=2184.048 Joules / 33.767g x 68.5°C

                                                                 c=0.944J/g*°C
                        ***Specific heat of metal in test 3 is 0.944J/g*°C

Test 4:   Finding Heat: q=89.23g x 4.184J/g*°C x 6.3°C

                                  q=2352.031 Joules

             Finding Specific Heat Of Metal: c=2352.031 Joules / 33.767g x 70.2°C

                                                                 c=0.992J/g*°C
                        ***Specific heat of metal in test 4 is 0.992J/g*°C

Conclusion: After comparing my calculated specific heat to the specific heat of the metals given to us, I have concluded that my metal was aluminum. Each of my calculated specific heats is within about 1/10th of the specific heat of aluminum, and one is withing 3 hundreths. each of the other metals are at least 4/10ths away from aluminum, and my calculated specific heats. Therefore, aluminum is the only metal that my sample could be.

Magnesium Metal and Hydrochloric Acid

Video: http://www.youtube.com/watch?v=Snm2F_YBTI0

A single replacement reaction where a metal replaces hydrogen in an acid.

Magnesiuym Metal + Hydrochloric Acid ---> Magnesium Chloride + Hydrogen

Mg_(s) + 2HCl_(aq) ---> Mg(Cl)_2(aq) + 2H_(g)

In this single replacement reaction, the magnesium metal is dropped into the hydrochloric acid. The magnesium dissolves in the aqueous and replaces the hydrogen. The hydrogen is released as a gas.

When the magnesium metal was added to the hydrochloric acid. The solution instantly started to sizzle around the magnesium. There was a visible gas given off and a slight amount of heat is created. When the reaction is over, the magnesium metal is no longer visible.

The amount of each compound used truly does not matter. But the more magnesium you use, the longer the reaction will last. To compensate for more magnesium, you must add more hydrochloric acid.

We were able to preform our reaction without any noticeable errors.

Learning Target

Determine whether a compound is ionic or covalent by simply looking at the formula.

This can be done very easily. Ionic compounds have one metal element, and one non-metal element. Covalent compounds are composed of 2, non-metal elements.

Ionic Examples:
ZnCl2, Zinc Chloride. Zinc is a metal, Chloride is a non-metal.
AgNO3, Silver Nitrate. Silver is a metal, Nitrate is a non-metal.

Covalent Examples:
SiF3, Silicon Trifluoride. Both are non-metals.
PO4, Phosphorus Tetroxide. Both are non-metals.