how to find initial rate in an enzyme catalysis experiment?
955797
Use the Arrhenius equation to predict the rate:\n\nhttp://en.wikipedia.org/wiki/Arrhenius_equation\n\nK = Ae -E/RT\n\nIt all boils down to your reactants, the activation of energy, and the product that is being synthesized.
149491
This is a "rule of thumb" that is the average of a lot of kinetic data.  To blindly apply it is unwise, but as a zero order estimate, it should work OK.  Detailed measurements must be done on your system of interest to find out true kinetic parameters and their sensitivity.
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
818036
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
163081
finding out ways to produce enzymes quickly and efficiently
891159
use the equation\nE(enthalpy change) = m(mass) c(specific heat) DeltaT(change in temperature)\n\ntemperature change is initial temperature minus final\nsince temperature increased then the temp change is negative\n\n-2.00C
485922
the minimum amount of energy required to start a reaction
317089
yes, reaction rates are based on several things, concentration and temperature are the most important, because of the water added to the reaction, it will be less "likely" that a vinegar molecule will collide with a baking soda molecule, because of this the reaction will go slower. the way to calculate this is K = [Products]/[reactants].  Where [] is the concentration of each solute present (in moles per liter), and K is the constant value for this perticular reaction, which can be found in a chemistry text book.
520742
The activity of an enzyme is tested by measuring how quickly it degrades the substrate or how quickly product is formed in an enzyme-catalyzed reaction.  This can be measured using spectrophotometry.\n\nHowever, you cannot compare these activities in samples with different protein concetrations because the increased enzyme activity may only be due to the fact that there are more enzymes contained in the sample.  Therefore, you divide the activity by the total amount of protein in the sample.  This way you can compare the activities of different samples and get an idea of how well the enzyme is functioning in different environments.\n\nAs for the calculations, just asking me to do them is giving me nightmares about my biochemsitry course I just completed.  If you already have the activity, you divide it by the total mass of protein present in the sample to get the specific activity.  If you are doing a purification procedure, the more you purify the sample, the less protein will be present and as long as the enzyme of interest remains in the assay samples the specific activity should go up.
572399
Well, the kinetics are actually pretty complicated. \n\nAll of the catalysts I'm aware of for this type of reaction are heterogeneous. This complicates things for several reasons. First, the rate is going to be much more dependent on reactor design since how well the catalyst is suspended and dispersed plays a big role in the rate of reaction. Second, the rate will depend on the active surface area of the catalyst, which can vary from batch to batch of catalyst and must be determined empirically for each sample.  \n\nAnother tricky issue is that in the course of the reaction you are going from a liquid (oil) to solid (fat). This will change the physical properties of your mixture, which changes the way the catalysts is distributed and will certainly change the kinetics. Solution viscosity could potentially become an issue, especially on the industrial scale. It really just depends on how you run the reaction.\n\nIn terms of amount of H2 needed, it's going to depend on precisely what kind of oil you are starting with. Most vegetable oils are mixtures of several oils made of the triacylgylcerides of unsaturated fatty acids such as oleic acid and linoleic acid. Since the degree of unsaturation varies from acid to acid (and therefore oil to oil), it depends entirely on which specific oil you have.\n\nIn terms of catalyst, in principle it's always regenerated and never lost.\n\nI wish I could give you more specifics, but the kinetics here are really messy and not easily amenable to a traditional analysis.
723534
http://en.wikipedia.org/wiki/Enzyme\n\nyour answer is in industrial applications
697545
1- find heat quantity\n Q=C*t\nQ = heat quantity of the reaction\nC= heat capacipy of calorimeter\nt=temperature increasing\n\n2- find heat of reaction \n H=(Q*1000)/(N*V)\nN = normality of acid and base\nV= volume of acid and base
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
268703
Use the overall rate equation: rate = k[A]^x[B]^y...\n1) define for rate1 and rate 2 with the appropriate concentrations.\n2) divide both sides (e.g. rate1/rate2 = ...\nNote that this cancels k\n3) Solve for x, y, ...  These are your respective orders.\n4) Add x, y, ...  This is your overall order.\n\nYou can get the value of k by:\n5) Plug in values of x, y, ... to either rate1 or rate2 equation.\n5) Solve for k.
44844
The Km value of an enzyme is a tricky thing. It's a bit more complicated, but let me explain.\nFor most enzymes, Km is directly related to the its affinity to the substrate. The higher the value of Km, the lower the affinity. Of course, the lower the Km, then the higher the affinity. You noticed I said "most"...for some, this isn't the case. If you're interested in why, I'd suggest a more advanced book on enyzme kinetics. \nAs far as how Km influences efficiency, rememeber that the equation for how efficient an enzyme is is Vmax/Km. So, to answer your question, if you have two enzymes with equal Vmax, the one with the higher Km will be less efficient that the one with a lower Km.
915109
http://www.biologymad.com/master.html?http://www.biologymad.com/Enzymes/enzymes.htm\n\nhttp://www.tvdsb.on.ca/westmin/science/sbi3a1/digest/enzymes.htm\n\nhttp://www.activescience-gsk.com/miniweb/content/enzymes/how_do.htm\n\nI use these websites with my students...it has some basic enzyme info.\n\nAn enzyme is active only when there is substrate present. So therefore, the rate of activity will increase when thre is "something to do!"