Autocatalysis ?? Exponential or sigmoid or linear ? See below :)?
144359
If you have the reaction A + B -> C + D, under pseudofirst order conditions where B is in excess assuming a rate law of d[A]/dt = k[A][B] = k(obs)[A], observation of [A] should be a first order exponentional over time.  You can confirm the first order exponentional by plotting ln([A]) vs. time, which should be linear.  If you have some sort of autocatalysis - d[A]/dt = k[A][B][C], then the plot of ln([A]) vs. time will deviate from the linear with a larger deviation as the reaction progresses.  Another possibility other than autocatalysis for the not strictly linear log plot could be a change in the reaction medium as the reaction progresses.  If one of your products - say C - significantly more or less polar than A, you might be changing the rate of your reaction which the changing solvent mixture.
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
141538
Read the references at the end of the article or talk to an industrial engineer.\n\nhttp://en.wikipedia.org/wiki/Linear_programming
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.
644818
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
391116
2Ca(s) + O2(g) -> 2CaO(s)
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
871220
The second one:\nAF + H --> AFM + TM\n\nActivation E: +10KJ/mol is moderate\nHeat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).\n\nSee the site below:
1125729
The index-of-reaction equation (i.r.e.): dF/dtZ = kZ(l - βF)m is compared with a generalized Prout-Tompkins equation (g.P.T.e.): dF/dt = kn(l - βF)y With the boundary condition F = 0, t = 0 the equations are equivalent for z = 2 or m = 0 or β ≤ 1 and the i.r.e, is closely approximated by the g.P.T.e. for other values of the parameters by use of the following approximation ∫(1- F)-m dF ≈ AF(1-BF)5 The parameter s is a function of m. Both A and B are close to unity for 1/3 < m < 3. The i.r.e. is recommended as being more useful than the g.P.T.e. because the i.r.e. has a better theoretical basis and because it is more readily integrated and fitted to experimental data.
731401
Correct.\nD. when delta G is positive the reaction is non-spontaneous
152635
You program the data in to a calculator (like a TI-83), graph, and model the data points using an exponential profile.  It will then give you the form of the equation that best fits your data points in the form of a*e^(b*x), which you can then graph over your data points.  In your case, "b" should be negative to indicate decay.
392823
No, it will merely lower the activation energy required by the reaction allowing it to proceed.  It won't actually make the reaction spontaneous.  Spontaneous implies that the reaction does not require any outside influence which would negate the need for a catalyst. ;-}
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
824183
just go 2 google search & write the non linear edition tutorial  \nn u will get nice tutorials
698585
If the Catalyst is inorganic the higher the temp the quicker it works, if the catalyst is organic(enzyme),it will reach a peak of activity,then further temp will destroy the catalyst
883446
Hi, it depends on particular organisms. Check the link below for basic mathematical formulas you may be able to use.\n\nA study is done in which the number of bacteria in a liquid culture is monitored over a period of several hours. Ten observations were made:\n\n\nTwo questions need to be answered:\n\nIs this data consistent with an exponential growth model: N = N0ekt, where \nN0 = the number of bacteria present initially (at t = 0)\nk = a growth rate constant\nN = the number of bacteria present at a given time t,\nand e is the fundamental mathematical constant forming the base for natural logarithms. \nIf the answer to the first question is "yes", then what are the values of N0, k, and t2 (the so-called doubling time) for this bacterial system?\nThe experimental measurements listed above indicate the number of bacteria are growing with time. If we plot the observed values of N (vertically) against the elapsed time, t, in hours, we get a graph (available on web page below).\n\nYou can draw a smooth curve through the experimental points as shown in this figure, and the points all seem to lie relatively close to this curve. Further, this curve looks quite a bit like the graph of an exponential function, but from what we see, it might just as well be part of a parabola or of one of a number of other curves that result when various algebraic forms are graphed. This graph suggests that the answer to the first question may be "yes", but we would usually want stronger evidence than this to draw a definite conclusion. The trouble here is that it is difficult to distinguish precisely and reliably between the curves are the graphs of quite different mathematical functions.\n\nThere is, of course, one curve that is easily recognized by everyone - the straight line! In this situation, the logarithmic function can be used to rewrite the exponential growth formula into a form that will allow us to create a plot that does give a straight line when data is the result of an exponential growth (or decay) process. Start with the exponential growth formula:\n\nN = N0 ekt\n\nSince this involves an exponential with base e, it is "natural" to use natural logarithms - those based on the number e. Using the more common notation ‘ln’ for ‘loge’, we simply take the logarithm of both sides of the previous equation:\n\nln N = ln(N0 ekt)\n\n= ln N0 + ln(ekt)\n\n= ln N0 + kt ln e\n\n= ln N0 + kt\n\nHere, we have used two of the three basic properties of logarithms to manipulate the right-hand side, namely that\n\nlog (AB) = log A + log B \n\nlog (Am) = m log A \n\nFurther, in the last step we took advantage of the fact that by choosing to work with natural logarithms, ln e = logee = 1. Thus, the exponential growth formula can be written in the logarithmic form\n\nln N = kt + ln N0\n\nIf the exponential growth formula is correct and we were to plot ln N along the vertical axis of our graph (rather than N itself), and t along the horizontal axis, then the equation of the plot we are preparing will have the generic form\n\nvertical quantity = constant x horizontal quantity + another constant \n\nBut this is just the equation of a straight line. The plot that results using the data given above is available on web site below.\n\nAlthough the values written on the vertical axis are values of N, this graph is ruled vertically so that the vertical distances are actually proportional to ln N. Since one scale is logarithmic (the vertical one) whereas the other graph scale, the horizontal one, is the ordinary uniform spacing type, this is called a ‘semilogarithmic plot’.\n\nAt any rate, the figure indicates that the points are scattered about a straight line, the scatter due to small random errors in measurement. Since the plot of ln N vs t gives a straight line, we have confirmed that\n\nln N = kt + ln N0 (equation *) \n\nis a valid formula for this data, and therefore, so is its equivalent, N = N0 ekt, valid for this data.
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P