i need the formulas on elongation of carbon nanotube due to heat and bending...
200216
Ouch! I think you will have to go to a journal article for this.\n\nI was offered a project on fullerines, however i chose education instead.\n\nThe reality is that fullerines (nanotubes being a fullerine) are such a modern area of study that there are only a few experts in the world. These in deep study and often still developing information.\n\nI suggest an inorganic chemistry journal (although fullerines are carbon, most of the work done is inorganic chemistry).\n\nI think Mitsubishi recently started developing large stocks of fullerines for laboratories to use so hopefully more information on them will blossom from the research.
156326
Since you have 3 points plot those(number of carbon in each compound vs molar enthalpy)\n\nSO ethanol would be 2, -0.215\n\nThen you could try drawing a line of best fit(assuming the relationship is linear) between the three and extending the line till it reaches 5 carbon
5447
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
661530
Ha ha :))\n\n\nlet's say x be the length of metal rod 1 in the final composite rod \n\n===> (31-x) is the length of matal 2 in composite rod \n\nalpha 1(coefficient of linear expansion for metal 1 )  = change in length /original length  *change in temp =6.35* 10.^-2/31*100\n=\n\nSimilary ,for metal 2\n\n\nalpha 2  = (3.20 * 10.^-2)/31*100\n\nFOR THE FINAL compsite  rod  \n\nx*alpha1 * change in temperature + (31-x)* alpha 2* change in temp = 5.55 * 10.^-2 \n\nchange in temp above eqn is same =100\n\n0.002x+(31-x)*0.0010=5.55*10.^-2\n\n0.001x=0.0245\n\nx= 24.500 cm of metal rod 1\n\n& (31-24.500) cm = 6.500 cm of metal rod 2\n\n\nTHE author below has made a mistake of leaving the orriginal length :\n\nThe eqn is:\n\ndl=alpha*original length*dT
532386
Type "linear coefficient of thermal expansion" and "volume coefficient of thermal expansion" into you search engine.\nFor spheres, you probably want to use volume.\n\nYou should find formulas that relate change in length (or volume) and deltaTemperature.\n\nPlug the two temperatures and the coefficient into the formula and you will have the change in size.
354175
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
412072
you could take a straw and heat to just before the melting point where it starts to get soft. you could also use a small gauge (meaning the wire is thick) and bend that. Then attach the tube worm to that so it looks like some of it is sticking out.
335710
View source of the that page.\nYou will see the formulas.
364217
E = cp*m * ∆T; you have cp (specific heat), m and ∆T.  Find how many moles 1kg of iron is, and subsitute that for the mass in the formula for cp.  You can get that information here http://en.wikipedia.org/wiki/Iron
29609
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
431774
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
39053
You can use Bernoulli equation:\np1/ρ + (u1)^2/2= p2/ρ + (u2)^2/2\n\n Where ρ is water density\n p1 : initial presser\n p2 : final presser\n u1 : initial velocity\n u2 : final velocity.\n\nnote that u2=u1*(A1/A2)\n\nwhere A1 and A2 are areas related to input and output of the tube.
511881
See the link\n\nfoi iron Z=26  this is the formula of the distribution of electrons on shells and subshells\n\n1s2 \n2s2 2p6\n3s2 3p6 3d6\n4s2
449778
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
265564
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
73537
1.75mol  /  2mol of C6H6 times specfic heat ( -98.0kJ)\n\n=  .875 * -98.0kJ = -85.75kJ