What is the final Celsius temperature of the mixture?
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
1102200
Pour each into a separate container and heat them both up to 100 degrees Celcius.  The one that boils is water.
274160
32 degrees F\nor \n0 degrees C
1140288
B. 93.3\n\nThis will help you:  http://www.onlineconversion.com/temperature.htm
131196
The melting point is 1068 kelvin and 795 celsius. The boiling point is 3633 kelvin and 3360 celsius.
170755
Specific heat 0.13 J/gK
175159
Solid\nCopper melts at 1083ºC
670047
Melting point 933.47 K\n(660.32 °C, 1220.58 °F)\n\nBoiling point 2792 K\n(2519 °C, 4566 °F)
889851
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.
959193
-273.15 degrees celsius
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let final temp be x\nHeat lost by 0.5gm h20 = mass*specific heat(a constant)*(98 -x)\n= 0.5 *4.2*98-x\n= 2.1(98-x)\nheat gained by 0.235kg water = 235*4.2*(x-20)\n=987(x-20)\nheat lost = heat gained\n2.1(98-x) = 987(x-20)\n205.8 - 2.1x = 987x - 19740\n989.1x  = 19945.8\nx = 20.165 degrees C\nfinal temp = 20.165 deg C
88715
Latent heat of ice is 334kJ/kg.\nThis means, to convert 1kg of ice at 0c to 1kg of water at 0c it would require 334kJ (or 334000 Joules). \nSo, to convert 0.3kg of ice to water at the same temperature would require 0.3*334 or about 100kJ.\nThe answer would be (A) above.
521266
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
816920
Call p1 = p(Cl2), p2 = p(O2)\n\np1 = 0.15 atm, p = p1 + p2 = 0.365 atm\n\n-> p2 = 0.365 - 0.15 = 0.215 atm\n\nV = 0.538 l\n\nT = 273 + 32 = 305 K\n\nR = 0.082 L*atm / K\n\nFor a mixture, all gases share the same V and T:\np2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol\n\nTherefore, the mass of O2 is:\n0.004625 * 32g = 0.148 g
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
860053
measure the cold water temp and take a large amout of cold water than hot water and mix them .\nuse the realation mhot (Thot - T2) = mcold(T2 - Tcold)\nwhere:\nmhot = amount of hot water , Thot = hot water temp\nmcold = amount of cold water , Tcold = cold water temp.\nT2 = the temp. of water after mix.\n(considering the specific heat of water constant at this range of temp.)
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To convert fahreheit temerature to celsius:\nsubtract 32\nmultiply by 5\ndivide by 9.\n\nI'll let you do the math.