I know the 'AM' is for 'acid metabolism' but I can't remember the 'C' for the life of me.
211820
Crassulacean acid metabolism (CAM) is a carbon fixation pathway in some photosynthetic plants. CAM is usually found in plants living under arid conditions, including those found in the desert (for example, cacti or pineapple). It is named after the plant family it was first discovered in, the Crassulaceae.
426688
This brings back dreadful memories! Glucose is the principal sugar in the biological world. Breakdown of glucose happens in two stages. The first stage is glycolosis, which consists of 10 successive steps in which the glucose, a 6-carbon molecule, is converted into two 3-carbon molecules (pyruvate) and two ATP molecules are synthesized from two ADP molecules and two P i. The second stage of glucose breakdown is called celular respiration, converts the pyruvate molecules into CO2 and H2O through 15 steps, some of which are coupled to the synthesis of many molecules of ATP from ADP +Pi. When pure glucose is ignited and burned, it is completely converted to CO2 and H2O with the release of heat. C6 H12 6O2---->6CO2 + 6H20. The breakdown of glucose in a cell leads to the same result. Six O2 molecules are needed to catabolize each molecule of glucose, and 6 molecules of CO2 and 6 molecules of H2O are the end products. For one mole (180 grams) 686 kilocalories are released. In a cell, the breakdown occurs stepwise, with some energy released at each step. By coupling certain of the steps to synthesis of ATP from ADP, energy is captured. Glycolysis carries the breakdown to pyruvate, during which some energy is transferred into a net gain of two ATP molecules. Glycolysis, however, taps very little of the free energy in glucose; about 93% of this energy is still present in the two pyruvate molecules left after glycolysis. This energy is released in stepwise fashion by conversion of pyruvate to CO2 and H2O in cellular respiration, with the generation of 36 ATP molecules. Cellular respiration occurs in two stages. In the first stage, the pyruvate is oxidized in a sequence of reactions to CO2 and H2O, and electrons are transferred to coenzymes. In the second stage, the passage of the electrons along an electron transport chain is coupled to the synthesis of ATP from ADP and Pi in a manner similar to the transfer of electrons and generation of ATP that occurs in phase 1 reactions of photosynthesis. Then you go on to the citric acid or Krebs cycle. I hope this helps. I remember drawing it out over and over to memorize it. The main thing is to remember as the process goes on what the byproducts are-ATP, etc.
476231
Hey, Blond fuss?  Good luck if U can't find help @ this late hour. Keep working on it.  Something from what U learned so far must b in your head - find it. No one else working on this problem awake?\n  I'm tired (12:35 a.m. here @ UMAss Amherst).  Had 1st-yr Chem 100 in 1978, not Chem 1111 !! Barely remember orbitals.  U do have books there, don't U?   Library maybe not open now as ours is.    \n Aces 2 computer. So U have awhile 2 figure it out.  Good luck.
351460
I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
466512
Here is what you need.\n\n2HF(aq) + Mg(s) ---> MgF2(s) + H2(g)
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Because 36 ATPs is complete oxidation of a glucose molecule, from glucose all the way down to CO2 and water.  This is called AEROBIC metabolism, meaning oxygen was required to accept electrons as the final step.\n\nUnder anaerobic conditions, where oxygen is not available (i.e. during a spring, not enough oxygen is available to all of your skeletal muscle cells), glucose is not fully catabolized to CO2, instead it is catabolized to lactic acid, which nets much less ATP than complete oxidation would.
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bcd\n\n\n\nSolution:\nCaCN2 + 3H2O ---> 2NH3 + CaCO3
599962
I'm assuming that C' means not-C, rather than another variable.\n\nABC+ABD+C'D\n= ABC + ABD(C+C') + C'D\n= ABC + ABCD + ABC'D + C'D\n= ABC(1+D) + C'D(AB+1)\n= 1ABC + 1C'D\n= ABC + C'D\n\nWrite out the truth table and you can easily see that this is the case.
491742
After glycolysis is finished 2 molecules of acetic acid are "loaded" into Kreb's on the enzyme Acetyl Co-A.  Co-A then returns to glycolysis in the cytoplasm to get more acetic acid.\n\nMean time the acetic acid adds to oxaloacetice acid (A 4 carbon structure) and gives a 6 carbon structure called citric acid. (2+4=6).\n\nThen CO2 and NADH2 is formed and we drop down to a 5 carbon structure called alpha ketgluteric acid and then lose CO2 and NADH2.  We also have created enough energy at this point to form an ATP and a low energy "hydrogen bus" FADH2.\n\nWhich brings us back to the 4 carbon structure oxaloaceitic acid, and we start all over again. \n\nThe purpose of Kreb's is to load what I call hydrogen busses (NADH2) because they are what unload electrons in the electron transport system which gives us a huge amount of ATP's (34).\n\nHope this helps.
516758
the fomula is:\n\nCxHy + (x+y/4)O2 ---> x CO2 + (y/2)H2O
140333
It starts with 0.\nhttp://staff.jccc.net/pdecell/chemistry/phscale.html\nhttp://www.epa.gov/acidrain/site_students/phscale.html
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A -produces two molecules of CO2\n\nnot C because it makes NADH, not NAD+\nnot D becaues the most ATP is produced during the electron transport chain
377523
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
488074
OK, first of all, those little numbers at the top right aren't exponents, they're charges, so you don't need to use the ^ symbol.  It should be written "HCl(g) + H2O(l) -> Cl- + H3O+" or "HCl(g) + H2O(l) -> Cl-1 + H3O+1".\n\nThat aside, HCl and H3O+ are Lewis acids.  In fact they're very strong acids by any definition.
587931
ask your chem prof or tutor ..it's hard to explain over the internet!
304650
i think its "C" but im not sure...could be an informal though! lol
310685
It's a polynomial. You've added no NaOH so far, so it's:\n\nDraw an ICE chart, where ICE is initial, change, equilibrium. Let x be the amount of BA\n\nBAH  <---> H+ + BA-\nI 1.0 M    0     0\nC  -x      +X    +X \nE 1.0-X    X      X\n\nI don't expect the formating to surivive, so the equation becomes, based on the law of mass action:\n\nX^2/1-x = 6.3*10^-5\n\nThere are two ways to proceed at this point. We can assume that the amount of benzoic acid that has diassociated is much larger than the amount that has not. If we assume this, we disregard the -x and figure out:\n\nx^2 = 6.3*10^-5  so x = 0.007937.\n\n-LOG (0.007937) = 2.1\n\nSay we don't make that assumption, then we have to solve the equation: \n\n\nX^2/1-x = 6.3*10^-5