Hello!!! how can i convert log cfu/ml to g/ml (in microbiology) Please, i really need it.?
93236
A cfu is a colony forming unit, meaning a viable cell.  You'd have to know the size and density or the mass of a given cell to know this.  What exactly is the problem?
425236
I've worked with flowmeters for 20+ years and have never heard of this... what medium are you trying to measure?
1138340
First, you need to multiply this number with the volume you are given in mL to obtain the mass because you are given a density measure (mass per volume). Then, divide by 10^6 because ug means microgram, or a millionth of a gram.
142959
Well, actually you can.  If you know the specific volume of the material for which you wish to convert units, you can use volumetric measurements to determine mass.  Here's how you do it:\n\n# of grams * (specific volume, in L/gram) = # of liters\n\nSince specific volume isn't a very standard measure, you will probably need to convert from a measure of density (often in units of grams/cm^3).  Here's a more complete formula:\n\n# of grams * 1000 / (density in grams/cm^3) = # of liters.
789189
1\n\nIn the future, this site is excellent.\n\nhttp://www.onlineconversion.com/volume.htm
703531
FW of NaCl is 58.44  then\n\n   58.44 gm in 1000 mL = 1 Molar\n\n   4.78  gm in 200 mL = ?\n\ndo cross multiplication\n\n(4.78 gm x 1000 mL x 1 M) / (58.44 gm x 200 mL) = 0.4089 M\n\n(gm/gm and mL/mL cacels out so you get answer in M)\n\n\nNow you are taking 28 mL of 0.4089 M solution and diluting it to make final volume of 100.  Use M1V1 = M2V2\n\n0.4089 M x 28 mL = ? x 100 mL\n\n\nThen  ? = (0.4089 M x 28 mL) / 100 mL    { mL/mL cancels out and you get answer in Molarity)\n\n\n? = 0.1145 Molar is your answer
779549
OK , i'll give it try.  we are assuming that the water is distilled because that affects the specific gravity.  grams per cc of h20.  4 oz. of water =1 cup......so 1oz of water =.0296 liters.......4 x that  = .1184 liters.....devide that by 1000 to get the mass or gram  value of 1.184 cubic centimeters = 1 cup of water...give me 10 pts for effort.!!  I think i messed up on the conversion but believe the answer is correct!
324860
pKa + pKb = pKw  so -log (2.1x10E-11) + pKb = 14\n\nSo pKb = 14- ( - log ( 2.1x10E-11 ) )\n\n= 14 - 10.68 = 3.32.... I swear I worked this exact question out when I was an undergrad....lazy students!  We didn't have the damned internet!  ( well we did but I didn't use it ----- go and learn the stuff!!!  Be a real chemist! )
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
958491
i think we can convert the log with subsrcipt 3 to natural log\nlogx (with subscript 3) = lnx/ln3\nso derivative is\n1/(xln3)\nhope this helps
773889
Suspected trick question.  The volumes included are irrelevant.  You know the molecular weight (66000 Da) and you know how many micromoles you need total in each solution.  What does it matter if the one micromole is in 1 ml or 10K gallons?  You know it only contains one micromole.  Divide the molecular weight by 10^6 and you're done.  (or you have not stated the problem correctly)
951234
Grams are mass\nKiloliter is volume.\n\nYou can't convert 1 to the other.
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.
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^
212141
http://www.nolo.com/definition.cfm/term/6FF94E70-3B77-4047-8983A27F34E1DF28
904268
just add 3  zeros or multiply by 1000 (which is the same as adding 2 zeros o_O)\n\nexample: 78g=78000mg
342773
Want to see something neat??\n\nType "19 gallons in cubic feet." into google.\n\nWorks for all kinds of conversions.