i want to know thw grade of material those melting point must be higher than 1550'C and thermal expention on heating between 1200-1300'C will be 50% more than the expention in En-31 on the same temp. Material must be economical and easily available in the market.
207875
at the market?  so you mean a food product? or some type of hardware/metal?  a can or glass jar?  I'm trying to understand blindly here?  give me some more info so I can help you more. (I also must convert your temperatures) but be as specific about what materials you refer to please? email yahoo aplacefortess or addme to live chat. good luck
532272
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
512701
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
155111
specific heat? that would be represented by Cp.\nHowever if you are using delta H then it would either be: delta H (fus) which is the enthalpy of fusion, used in problems concerning melting point (indicating a state change from solid to liquid)\nOr it could be delta H (vap) which is the enthalpy of vaporization, used in problems concerning a boiling point (indicating a state change from liquid to gas)\n\nHeat calculations are done using the formulas:\nq=m x (delta T) Cp <-- Used when there is a temperature change, but no state change\ng= m x (delta H fus) <-- Used for state changes from solid to liquid\nq= m x (delta h vap) <-- Used for state changes from liquid to gas
16139
http://www.itt.edu/staff/boyter/links/index.html\n\nLook under chemistry equipment.  There are some chemicals you will not be able to get.  Posting what you are looking for would help.
619685
From a "Dangerous Goods" standpoint :\n\nFlammable materials have a flash point below 61°C. So, most common solvents (Stoddard solvent, kerosine, petrol, VM&P Naptha) and organic liquids and gases (acetone, ether, nitromethane, methane, ethane, propane, butane, ... up to about nonane) are flammable. Some solids (eg naphthalene) are also flammable.\n\nCombustible materials have a flash point greater than 61°C. So diesel fuel, paper, coal, wood, etc are all combustible.\n\nAlso, don't get confused between explosivity and flammability - many explosives aren't flammable. C4 is an example - it will burn (combust) but requires percussive detonation to make it explode. So here's a tip - don't try to "stamp out" burning C4!\n\nAlso, explosivity may be limited to a certain range of concentrations (expressed as Lower Explosive Limit - LEL and Upper Explosive Limit - UEL). These are usually expressed as the % w/v of the fuel in air, although it is sometimes expressed in relation to the oxygen content.
415802
Hi Kitty,\nHere is what, I think, you need.\n\nExothermic, dH < 0, Entropy increases (dS > 0), Reaction is spontaneous and dG < 0\n\nExothermic, dH < 0, Entropy decreases (dS < 0), Reaction is spontaneous only at low temps, if |T dS| < |dH|\n\nEndothermic, dH > 0, Entropy increases (dS > 0), Reaction is spontaneous only  at high temps, if T dS > dH\n\nEndothermic, dH > 0, Entropy decreases (dS < 0), Reaction is not spontaneous, dG > 0
414446
Lead — Specific Heat Capacity: 129 J/(kg*K)\n \n((129 J) / (kg * K)) * (.72 kg) * (7 K) = 650.16 joules
97902
no, reaction times are too slow at that temperature.  (I'm just guessing, but I'm a chemist.  You should be able to look up reaction energies, try Beilstein.)
370161
You use the formal \nH= mcdeltat\n\nmanipulate it to be...\n\nc (specific heat capactiy) = H /(mdelta t)\n\nChange temperature to Kelvin,....4degrees = 277Kelvin\n\nC=221/(55X277)\n\n\nC= 221/15235\n\nC=0.0145 (in JOULES)
151001
C whether this helps You,\nhttp://web.umr.edu/~kpeaslee/classes/Met221/KeyHW2.pdf#search='measuring%20SDAS%20in%20Aluminum%20alloy%20casting'\n\nhttp://www.aluplanet.com/eng/info_eco_doc_PO.asp?Mnu=141&Itm=163&Grp=0&Tgl=162&Doc=3804
488379
Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
483176
Your unknown compound might be an impure substance.  The impurities will skew the melting point.  You should find out if the substance is pure or not.  Pure substances (that is, with nothing else mixed with them) will have a narrow ranging in which they melt, for example, pure Naphthalene melts between 83-85 Celsius.
387056
Everything has a melting point and a boiling point.  In the case of water, the melting point is 0 C and the boiling point is 100 C.  In argon, as you have pointed out, the melting point is much, much lower, whereas for something like iron, it is 1535 C.  Neither of these temperatures occur normally on Earth's surface.  That's why you don't see puddles of argon lying around on the street, or clouds of iron vapor floating in the air:)
289042
Lava is molten rock. It is red hot which shows it is about 600 centigrade.\nThe melting point must be lower. It depends strongly about the kind of rock (basalt, granite, gneis, ...).
469283
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:
268054
You need the specific heat of water, which is 4.184 J/g*C.  Assuming a density of water of 1 g/ml (not always the safest assumption, but close enough), then your 500 ml = 500 g.  You're raising the temperature 35 degrees, so the heat you will need will be:\n\n4.184 * 500 * 35 = 73,220 J = 73.22 kJ.\n\nThe combustion of decane provides 6.78 MJ/mol, or 6780 kJ/mol.  In order to give 73.22 kJ, you would need to burn 73.22/6780 = 0.0108 mol of decane.  Given decane's molecular weight of 132 g/mol, you would have to burn 1.425 grams.