how to draw condense formula for 1,2-dichlorocyclobutane?
347783
Answer is at question 1. E.
390546
I'll give you a hint.\n\nDelta G = Delta H - T Delta S\n\nUsing that formula, you can get 1\n\n(B) you can look up in any physical chemistry text:\n\nG(2) - G(1) = a formula I don't have available
770269
For aA+bB -> cC+dD\n\nE=E0-(RT/nF)* ln([C]^c*[D]^d / ([A]^a*[B]^b))\n\nwhere n=number of electrons taking part in the reaction\n\nSolids are not considered in the equation, so in your case\n\nE=E0 -(RT/nF)*ln([Fe2+]/[Cd+2])\n\nYou can find the standard reduction potentials for the Fe+2 and Cd+2 at \nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\nThen E0= E0(Cd2+)-E0(Fe2+)\nR=8.314\nn=2\nF=96500\n\nYou have to specify T ...
773838
The formula would be 2HCL+ CaCO3 ---> CaCl2 + H2O + CO2. The standard enthalpies of formation are -167.2, -1207, -796, -285.8, and -393.5 respectively. When you use the formula "Sum of products minus sum of reactants, you come up with +66.1 (don't forget to multiply the HCl formation by 2 for 2 moles). I am assuming that the HCl you are talking about is HCl (aq) and not HCl (g). If it is a gas, then the standard enthalpy of formation is -92.31, so you would come up with a different number (-83.68)\n\nIf it is aqueous HCl, you come up with a positive 66.1 kJ/mole, which means that heat was absorbed by the reaction (endothermic). That probably means your cylinder would feel cooler, since heat is being taken in by the reaciton.\n\nIf you mean gaseous HCl, you come up with a negative 83.68 kJ/mol, which means that heat was given off by the reaction (exothermic). That probably means yoru cylinder woudl feel warmer, since heat is being given off by the reaciton and taken in by the cylinder.
875835
OD=E*d*C\n\nWhere OD=optical density\nE the molar extinction coefficient\nd the length of the path of the light through the solution (width of the cuvette)\nC the concentration in mole/L of the solute.\n\nYou need d and C as well to get E.\nThis is a fair approximation especially for student problems.\n\nIn reality if you wanted to find E, you would do a standard curve by measuring the OD of samples of known concentration.\nThen you would use linear regression to find the equation of the curve and from its slope you would calculate E (slope=E*d)\n\nA good standard curve and thus accurate calculation requires care to ensure the accuracy of the standards, several points (ideally 9 at least) and being within the linear are of the method.\nThe region of highest accuracy is 0.2-0.8. You can go even below or above this range (though not above 1.0) but this is the range where the photometric error is minimal.
63905
This may not be entirely correct, but I'm fairly certain that 2, 2 dichloropentane would be a five carbon hydrocarbon. The second carbon (from left or right, it doesn't matter) would have two chlorines attached to it, rather than two hydrogens. \n\nSo, this structure would be: CH3 - CCl2 - CH2 - CH2 - CH3.
93462
I hope this helps:\n\nhttp://chemistryformulas.com/
635418
CH3CH2CH2CH3 + CL2 -hV--> \nCH3CH2(CL)CHCH3 + HCL ..\n\nwhat are the ways to get the CL at the end of the chain..so that\n\nCH3CH2CH2CH2CL + 2NA + CLCH2CH2CH2CH3\n--->CH3--------CH3\n\nThat's a radical reaction, the chlorination of an alkane with chlorine in UV light. The intermediate is a carbon radical and secondary radicals are more stable than primary. There are also more secondary carbons on this molecule. So probability and reaction kinetics dictates that the primary alkyl halide will be a minor product.\n\nI guess, if this were my synthesis, I'd:\n\n1. Put the glucose in HCl to cleave the anomers it likes to make.\n\n2. Esterify the carboxlic acid with whatever's available (Fisher esterification with MeOH, for example)\n\n3. Reduce the ester with Diisobutyl aluminium hydride (DIBALH)\n\n4. React the aldehyde with 1 equivalent of Ethyl Magnesium Bromide (Grignard)\n\n5. Reduce the hydroxy groups with a Barton-McCombie deoxygenation, although I've never seen it done in a case like this.\n\nThen again:\n\nI'd go to the stockroom and save my research advisor a lot of money. :)
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
980626
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
469836
(a) 1,2-dibromobut-1-ene\n(b) 2,2-Dichlorobutane (based on Markovnikovs rule)\n(c) don't know\n(d) don't know.  Na is often used to reduce multiple bonds in a free-radical manner, however you need a source of H.\n(e) KMnO4 is a very strong oxidising agent and is commonly used as tests for unsaturation.  The side product is MnO2, showing that two atoms of oxygen are transferred to the product.  The product formed I think is 1-hydroxy-butan-2-one.
868179
2Rb + 2 H20 --> 2RbOH + H2
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
319800
Ok.  Butanal is CH3 - CH2 - CH2 - CH = O.  According to naming scheme, you would have the CH being 1.  Therefore, it is \n\n\nCH3 - CH2 - C(CH3)2 - CH = O\n\nSince I cannot draw it, at the C(CH3) is supposed to signify that on the C there are 2 methyl groups attached.
644051
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
956604
Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
728797
To be perfectly exact, the melting point of the alpha modification, which has a monoclinic crystal structure, is 53.5 C, the beta form, which is triclinic, has an mp of 54 C.\n\nI suspect your p-dichlorobenzene was contaminated, perhaps with water, causing a freezing point depression.  Another possiblility is that the sample cooled so rapidly, that you overshot the freezing point.