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Gardening: Interlude

During my break between the Spring and Summer terms I’ve been spending a lot of time in the garden. It is different every day. There are challenges – like when you plant broccoli and slugs eat the young plants or when spider mites destory the jasmine follage (both of these are being battled with soapy water) but there are also so many rewards.

Hydroponics Planter

Hydroponics Planter

Paolo built a hydroponics system to grow strawberries. It is amazing. It uses a bell siphon to drain the water and fill the trough over and over. The strawberries are planted in a low density clay medium and fed from the goldfish pond. The flowers that are visible in this photo are: roses, clematis, columbine. begonias, and coral bells.

Over the last three weeks some flowers have already come and gone. The tulips are gone and I miss them. They were spectacular. But today I saw my first primerose of the season and a few poppies have greated me every morning.



I can hardly wait for the rasberries. They should begin to appear in June. All of that green, green folliage on the right is rasberry canes. A cane takes two years to bear fruit then it dies. The secret to heathy rasberry plants is to prune the old canes each year. The wisteria forms the purple cannopy over the pond. Can you spot the fennel? It is the light green fern plant on the left of the columbine.

Roses and Lettuce

Roses and Lettuce

Today I cut some lettuce leaves for a fresh salad. I am growing lettuce in a window box. I keep moving it around the yard looking for the best spot. Lettuce does not like to be too hot but it needs lots of sun to grow. There is another giant sunflower in the making for this year!

Recently we visited the Botanical Garderns. What an inspiring place! I started a new bonsai plant (little container) when I came home. I used a maple tree that was trying to grow in the yard. The other bonsai plant is a juniper. It is about four years old.

Yard in Late May

Yard in Late May

This photograph gives you a bird’s eye view of the backyard. Borage is growing all around the rain barrel. Lemon grass is growing in the orange pot. The raised bed has the broccoli – but it also has slugs (yuck). We’ve been putting out containers of beer. It attracks the slugs and they hop in and drown. You can just see the tomato plants in the greenhouse. They’ve grow so fast. Compare what you see to the photo below that was taken in March. Notice the tulips and daffodils – they are all done now.

Yard in Late March

Yard in Late March

So I have to say it is pretty exciting to see so much change so fast. It is a lot of work to develop a garden but it is also incredibly satisfying.

PS 107: Plastics (Student Video)

Patrick Erwin made this video about Plastics:

Chemistry 201: Carbon Monoxide

On the last exam you were asked to consider carbon monoxide lewis dot structures and to comment on them. This excerpt copied here from Wikipedia does a good job of discussing the main points of this question:

Resonance structures and oxidation state

picture of carbon dioxide
Resonance structures of carbon monoxide
Different (correct) Lewis structures can be drawn for carbon monoxide. In the structure with three covalent bonds, the octet rule is satisfied, but the electropositive carbon has a negative formal charge. The structure with two covalent bonds would be consistent with the very low dipole moment of the molecule if the bonds were nonpolar. The structure with one covalent bond expresses the greater electronegativity of oxygen and the calculated net atomic charges. None of them do exactly meet the real electronic structure. Calculations with natural bond orbitals show that the structure with a triple bond is the most important Lewis structure (for the free molecule); this structure is the best approximation of the real distribution of electron density, with maximal occupation of bonding orbitals and lone pair orbitals.This is in accordance with other theoretical and experimental studies that show that, despite the greater electronegativity of oxygen, the dipole moment points from the more-negative carbon end to the more-positive oxygen end. The three bonds, however, are in fact polar covalent bonds that are strongly polarized. The calculated polarization toward the oxygen atom is 71 % for the σ-bond and 77 % for both π-bonds The oxidation state of carbon in carbon monoxide is +2 in each of these structures. It is calculated by counting all the bonding electrons as belonging to the more electronegative oxygen. Only the two non-bonding electrons on carbon are assigned to carbon. In this count, carbon then has only two valence electrons in the molecule compared to four in the free atom.

Chemistry 201: Hydrogen bonding

Three requirements exist for hydrogen bonding to occur. The first is rather obvious – hydrogen must be present. The second requirement is that hydrogen must be bonded to a highly electronegative atom and in actually fact only three atoms suffice: nitrogen, oxygen and fluorine. It isn’t only the electronegativity that is important but the intensity of the dipole that forms in the molecule. Thus even though chlorine has a high electronegativity it does not withdraw the electron density from hydrogen sufficiently for hydrogen bonding to take place. This is partially due to its larger size and additional shell of electrons shielding its nucleus. The final requirement is that there must be a non-bonding (lone) pair of electrons available and it is this lone pair that forms a strong interaction with the hydrogen nucleus. So to summarize:

  1. hydrogen must be present
  2. hydrogen must be bonded to nitrogen, oxygen or fluorine
  3. a lone pair of electrons must be present

Hydrogen bonding can occur between the same molecules or between different molecules (in mixtures) or within a long chain such as exists in carbohydrates, proteins and DNA.

Chemistry 201: Heating/Cooling Curve Calculations

Here are two problems from the last exam:
Diethyl ether, used as a solvent for extraction of organic compounds from aqueous solutions, has a high vapor pressure which makes it a potential fire hazard in laboratories in which it is used. How much energy is released when 100.0 g are cooled from 53.0°C  to 10.0°C?
Boiling point: 34.5°C    Heat of vaporization: 351 J/g
Specific heat capacity, (CH3)2O(l): 3.74 J/(g ∙ K)
Specific heat capacity, (CH3)2O(g): 2.35  J/(g ∙ K)
a. 10.1 kJ             b. 13.1 kJ               c. 16.1 kJ             d. 45.2 kJ               e. 48.6 kJ

Hint: ΔT X mass X S.H. = energy

(53-34.5)(100.0)(2.35) +(100)(351)+(34.5-10.0)(100.0)(3.74) = 48600 J

Liquid  ammonia (boiling point = -33.4°C)  can be used as a refrigerant and heat transfer fluid. How much energy is  needed to heat 25.0 g of NH3(l) from -65.0°C to -12.0°C?
Heat of vaporization: 23.5 kJ/mol
Specific heat capacity, NH3(l): 4.7 J/(g ∙K)
Specific heat capacity, NH3(g): 2.2 J/(g ∙ K)
a. 5.5 kJ   b. 6.3kJ c. 39 kJ d. 340 kJ                 e. 590 kJ
In both of these problems the key to solving them is in recognizing that there are three separate caculations that must be done. A substances is heated or cooled until it reaches a phase transition temperature. It then undergoes a phase transition. Finally it is heated or cooled. Use the specific heat capacities when temperature changes and use the Heat of vaporization during a phase change. Calculate the energy for each process then add the three energies together for a total energy.

Chemistry 201: Final Exam

Today I will post a few answers from the third exam. Also I want to make a clear statement that the exam will not cover colligative properties but it WILL COVER solution concentrations.

Why does double and triple bonding occur primarily with elements in the second row of the periodic table?

SIZE MATTERS. The main reason multiple bonding does not occur in the third row and beyond of the periodic table is because of the structure of multiple bonds. Multiple bonds are formed from UNHYBRIDIZED orbitals. They occur when a single electron is left in two p-orbitals for example and those p-orbitals overlab sideways. With big atoms the overlap is inefficient because the distance between the two nuclei is too large. ALSO, atoms in the third period and beyond have another option – expanded valence. So there are ways to form single bonds to more atoms using d-orbitals. This is the SAME reason H2Si2 doesn’t form. Silicon is too large to form pi bonds with itself. Some third row atoms do form double bonds with oxygen but that is due to the smaller nucleus of oxygen allowing the atoms to approach each other closely enough for sufficient p-orbital overlap. It is true that silicon is not as electronegative as carbon but that is not the reason why H2Si2 doesn’t exist. In this case it is simply the size of the atom that matters.


More in my next post…

Physical Science 107: Jeopardy

The latest jeopardy game, jeopardy fire, is now posted:

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