How do you find the rate of decay and the decay factor from data...time vs temp...time increases and temp decr
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.
251198
id say reactions can occur at any temperature above 0k. so 0.00000000000000000000000000000001K, the reaction would happen but really really really really relaly slowly. but thatz in theory. i have no clue bout pracitcally. i think the lowest temp theyve got to so far is either 4K or 0.6K. so ya.
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
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
608010
epsilon=0.6\nsigma=5.6704*10^-8 [J/(s*m^2*K^4)]\nP=Q/A=120 [W/m^2]\n\nWe know:\n\nP=epsilon*sigma*T^4\n\n120=0.6*5.6704*10^-8*T^4\n\nSolving for T we get:\n\nT=(120/(0.6*5.6704*10^-8))^(0.25) = 243.7 [K]\n\nThen, using Newtons cooling law:\n\nT(t)=T(environment)+(T(initial)-T(environment))*e^(-rt)\nT(t)=293+(243.7-293)*e^(-r*t)\nT(t)=293+(-49.3)*e^(-r*t)\n\nNow, I'm stuck, because I don't know the rate of change of temperature (r).  Maybe you try a different formula!?
926537
The decay is of the form N(t)/N0 = e^-t/h, where h is a constant.  When t = HL (half life time) N(t)/N0 = .5  therefore N(HL)/N0 = .5 = e^-HL/h.  Take the natural log of both sides to get ln(.5) = -HL/h, so h = HL/ln(2).  Therefore the formula for decay is \n\nN(t)/N0 = e^-t*ln(2)/HL\n\nYou want to find tx such that N(t)/N0 = 100/10^6 = 10^-4.  That means 10^-4 = e^-tx*ln(2)/HL.  Again, taking log (base 10) of both sides gives you -4 = log(e^-tx*ln(2)/HL) or[ -tx*ln(2)/HL]*ln(10).  The time tx is then 4*HL*ln(10)/ln(2).  This comes out 74,411years
930232
the time it takes for half of the material to decay
608949
cant answer secifically, but you should be able to back calculte taking elements present today and using the half lives as your main variable.
1187086
Higher temp = faster diffusion. Higher temps give the atoms more heat, which is basically a faster rate of the atoms vibrating. If they move around faster, they'll obviously diffuse farther in the same amount of time. Next time, do your own homework
212309
The half-lives will vary depending on the isotope.  Go to the library and look them up in "The Handbook of Chemistry and Physics" in the reference section.
885608
Using a simple TI-83, the data in #3 gives a linear regression equation of\ny = 0.4668x - 31.737, where y is absences and x is temperature.  r = 0.977.\n\nSound like you've been partying too much and ignoring your classes.  Well, you get what you choose.
266476
it has gone thru 2 half lives...after the first it emits 20 beta per min while on completion of the 2nd its emission rate becomes 10.. so its age would be 2 x halflife(C-14).which is equal to 11460 yrs.
676662
What did you use to measure the activation energy?\nYou probably used a thermometer which was probably marked to 0.1 degrees. That means that your measurements are probably accurate to 0.05 degrees. Reactant temp may be measured as 25.1 (+/- 0.05) so this means that the temp could be 25.05 or 25.15.\nDo the same for the transition state.\nDraw small vertical bars through each point, e.g for reactants the bar would go from 25.05 to 25.15, with the cross in the middle.\n\nI can't remember the formula, but errors are a lot more pronounced on small measurements. It may be something like the difference between the highest and lowest possible values divided by the recorded value, all multiplied by 100.
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.
22652
In 1.6 s, 1.2 g decays to 0.6 g\nIn 1.6+1.6 s, 0.6 g decays to 0.3 g\nIn 1.6+1.6+1.6s, 0.3 g decays to 0.15 g\nin 1.6+1.6+1.6+1.6 s, 0.15 g decays to 0.0750g\n\nSo after 6.4 secs, 0.0750 g remains unchanged
860007
5 Half-Lives.\n\nAfter 1: 200ng\nAfter 2: 100ng\nAfter 3: 50ng\nAfter 4: 25ng\nAfter 5: 12.5ng
1066565
Although some of the answers you have received are accurate they aren’t exactly succinct or to the point.\nYou haven’t however said what age the kids are you need to explain it to, please forgive me if this explanation seems too simplistic.\nThe theory of radioactive dating works along these lines.\n\n1. SOME ELEMENTS “DECAY” (CHANGE) FROM ONE THING TO ANOTHER, e.g. from Potassium to Argon or from one isotope to another e.g. C14 to C12\n2. THIS RATE OF CHANGE IS KNOWN, the speed at which it changes is known as the half life, i.e. how long before half of the total has changed\n3. A COMPARISON IS MADE i.e. you work out how much there was and how much has changed\n4. THE AGE IS CALCULATED. Once you know how much has decayed you can compare the amounts on well-known charts e.g. if you know halve the amount of C14 has changed to C12 then the sample is 5730 years old.\n\nIt is considered accurate because the rate of radioactive decay is absolute, measurable and observable. It has been tested numerous times and can be experimented on. The same results can be achieved and repeated by numerous people at many different labs. On the whole the decayed products (e.g. C12 etc.) are not found naturally. The original material is created at the time the sample is formed or moved (e.g. when the plant, animal grew).\n\nHowever the problems are\n1. In order to calculate the percentage of element changed you have to know how much there was. Sometimes the sample can have some of the original material removed or more of the original material can be added, e.g. new living fungi can grow and die on a log hundreds of years after the log was originally felled and buried.( this would give a younger date)\n2. The resulting product can be removed or further altered e.g. Argon from potassium can escape as it is a gas ( this would give you a later date because it would seem not as much or the argon had been formed)\n3. This time clock can be reset due to outside influences. A fire can reset the amount of C14/C12 because the carbon from the fire is different from the carbon formed in the living thing (e.g. the problems with dating the shroud of Turin)\n\nThis of coarse assumes that the equipment used is accurate and that no contamination was added e.g. someone touching or sneezing on the sample.\n\nOn the whole though in theory the method should be extremely accurate. Given perfect samples, good laboratory practice and accurate measurements the results seem impressive. This shouldn’t automatically however be taken for granted.\n\nThe final thing to remember though is that there is no other way of giving empirical, measurable dates on extremely ancient rocks or artefacts.