what is the heat capacity of liquid sulfur?
551718
GIVE ME POINTS LOOK IT UP YOURSELF LAZY GIRL
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
84471
Higher specific heat capacity liquids can absorb more heat energy per unit of liquid than lower specific heat liquids.\n\nIn most applications the higher specific heat liquid will be the liquid of choice as it can "carry away" more heat per unit volume.
706503
find the heat evolved, assuming heat cap of water is 4.2J/g/K\n\n(31.43-25.32) x 4.2 x 100g = 2566.2J\n\nnow according to definition, molar heat of neutralisation is calculated upon the mols of water formed. this number of mols is equivalent to the no. of mols of the limiting reagent (in this case, the HCl).\n\nmol of HCl= 0.050 x 1 = 0.05mol\n\nenthalpy of neutralisation= 2566.2/0.05= 51324J/mol\n\nthis answer is pretty close to real-life experimental values for strong acid-strong base.
1019766
sulphur exhibits valencies of 2 , 4 or 6 depending upon the conditions of reaction...it exhibits +4 oxidation state in SO2,+6 in SO3,H2SO4 and so on+2 in S2Cl2...is tthat enough...
834399
Your answer depends on the nature of your problem, so you'll want to give more info if you want a complete answer.  The specific heat capacity you use depends on which substance is being heated up or cooled off.  The mass will be the mass of the substance that is being heated up or cooled off.\n\nOne exception that frequently occurs is when you are dissolving a substance in water.  Most problems will assume that the solution has the density and specific heat capacity of pure water.  So in other words, if you dissolve 3 grams of ammonium chloride in 100 mL of water and watch the temperature go down, most instructors would want you to use a mass of 100 grams (instead of 103) and the specific heat capacity of water 1.00 cal/g C or 4.184 J/g C if you're working in joules.\n\nHope this helps...this is the best I can do without more info about the problem.
1025721
copper I sulfide---sulfur has a charge of -2, so you have to have 2 moles of copper with a charge of +1 for every mole of sulfur.  Copper is one of the elements that has a variable charge--but copper is always +1 or +2, you can tell like that by the charge of the other element
360214
Regarding the heat capacity of neon, I was able to find the following information along with some other facts about neon at Dr. John Grohol's site, "Psych Central."\n\nNeon\nHeat Capacity: (25 °C) 20.786 J/(mol·K)\n\nIncluded below is the source link for this information, for more data about neon.\n\nHope this helps!
21495
the formula is just the heat of fusion for water (in J/kg) times the amount of water you have.\n\nThe heat of fusion in Joules per kg for water is 333548 J/kg, so you just multiply that by the mass in kg to get the joules, so you'd have 333548 J/kg x 40kg = 13341920J which, in scientific notation would be 1.33 x 10^7 J.
1186743
This question would have been much easier if the specific heat capacity of coal were known!\n\nNevertheless, let us model the coal to be anthracite of composition 85% graphite and rest 15% water. \nSo the s.h.c = 0.85*0.17 + 0.15*1 = 0.2945 cal/g-°C\nwhich gives heat required as 0.2945*(220-20)*5000 = 294500 cal\n=294500*4.184=1,232,188 J\nSo it seems the answer is C\n\nTo check this we can back calculate the s.h.c(specific heat capacity) using ans C which gives the graphite content in coal as 78.8% (quite realistic)\nUsing ans D the s.h.c is greater than for water(not physically possible) \nAnd A and B can be eliminated simply by order of magnitude analysis.\n\nIt seems most likely that C is the ans.
211914
It depends on the mass. The latent heat of fusion of water (the energy required to freeze it) is 333.7 kJ/kg. That means to freeze one kilogram of water, it requires removing 333.7 kilojoules of energy.\n\nThis assumes that the water is already at freezing temperature. If not, energy must first be removed to get it at freezing temp.
673582
This information should enable you to solve the problem. \n\nLatent Heat of Freezing or Solidification \n\nThe amount of heat required per unit mass of a substance at freezing point (constant) and under 1 Atmosphere pressure to convert it from the liquid to the solid state is called Latent Heat of Freezing.\n\n \ne.g. At 00C & 1 Atmosphere pressure 80 K-calorie of heat supplied to 1 Kg of ice melts it into water and 80 K-Calorie of heat taken away from 1 Kg of water converts it into ice. \n\n&#92; L = 80 Kcal / Kg = Latent heat of fusion of ice\n\n\n    Latent heat of freezing of water \n\nso if 80 Kcal/Kg= latent heat of fusion of ice\nthen you can cross out the K and you have 80 cal/gram is the heat of fusion of water. but you hve 25 grams of water to freeze, not just 1 gram. so what comes next. what do you have to multiply by 25?\n80 right? \nthats right. see it wasnt so hard was it?\nso the answer is 2000calories or 2 Kcal.
92881
Maybe this will help (Lexan LS2)\n\nCoefficient of linear thermal expansion (parallel) \n0.7 E-4/°C
43501
Its a Metal, hence responds to / is a conductor of heat.\n\nIts fluid so the expansion per gradient of heat would be highest.\n\n\nAll other liquids are not a conductor of heat ( or as good as mercury)
855413
Latent heat of fusion of ice is only 80 calories/gm whereas the latent heat of vaporisation of water is as high s 536 cal/gm.Therefore C is correct
143592
Heat required to rise the temperature of  ice from \n-10 degree centigrade to 0 degree centigrade \n= ms(t2-t1)\n=10 * 2.1 * 10  g* {joule/(g*degree)} (degree)\n=210 joule\n\nHeat required to convert ice to water  \n=mass x specific latent heat fusion of ice \n=10 * 336  g*joules per gram\n=3360 joule\n\nHeat required to rise the temperature of  water from \n0 degree centigrade to 100 degree centigrade\n=ms(t2-t1)\n=10 * 4.2 * 100\n=4200 joule\n\nHeat required to convert water to steam\nmass x specific latent heat of vaporisation of water\n10 * 2264\n22640 joule\n \nHeat required to rise the temperature of steam from \n100 degree centigrade to 110 degree centigrade \n= ms(t2-t1)\n=10 * 2 *  (110-100)\n=200 joule.\n\ntotal of all is  30610 joule.
245442
Wait...not mixed with H2O? As in, molten liquid ammonia and molten liquid sodium sulfate? In that case no reaction. But with water, then:\n\n2NH3 + Na2SO4 + 2H2O --> (NH4)2SO4 + 2NaOH