.When a heat of Q is supplied to a diatomic gas in a certain process it does a work of Q/4 . Molar heat capacity of the gas in that process I \na. 2R/5b. 5R/2c. 10R/3d. 6R/7
209971
'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3
553353
You need to find the heats of formation of each species (will come from a standard table) and use this equation to get the heat of reaction:\n\n(delta)H = sum of heat of formation of products - sum of heat of formation of reactants\n\nEach heat of formation will be multiplied by its stoichiometric ratio, and the physical state will be important (especially that of water, as the heat of formation of steam is higher than that of liquid water).  You will not find a heat of formation for oxygen gas, because by definition there is no heat of formation for an element.
368706
First calculate the number of moles of acid and base that react by multiplying the molarity by the liters of solution.  The total volume (mass) of water is the sum of the two volumes (masses).  The calories attributed to the water are calculated by the change in temperature times the mass of water times the specific heat of water.  The calorimeter soaks up 480 J/degree times the temperature change, which is pretty stout.  (I am separating the two heat capacities.  If the 480 figure is for the water plus the container, then just use 480 times the temperature change.)  Then scale the answer up to one mole of reactant.
576603
like u said, if Keq = ([A]^a [B]^b)/([C]^c [D]^d), if a+b not equal to c+d, Keq will have units, OR will it?\n\nequilibrium constant has no units is because it  is always written as the ratio of the chemical ACTIVITIES of the products to the chemical ACTIVITIES of the reactants.\n\nchemical activity has no units, it is not but can be approximated to be the concentration of the soln for a soln(normally, a = C/ 1M, a = activity, c = concentration). The activity of a certain gas has no units also because of this very reason. We divide the pressure of the gas with standard pressure.
64923
Q=mc(Tf- Ti)\nWhere Q is the heat added\nm is the mass\nc is the specific heat capacity\nTf is the final temperature\nTi is the initial temperature
425847
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
14010
Find the values of heat of formation (delta H)of each molecule and follow this equation\nyour answer = Ca(OH)2 + H2 - Ca - 2(H2)
241242
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
497865
The change in energy is change in heat + change in work (dE = q + w)\n\nIn this situation, the system is releasing heat to the surroundings, , so it is exothermic and q < 0.  q = -350 J.\n\nIn this situation, the system is doing work on the surroundings, w < 0.  To calculate work, W = PdV.  In PV work, since J = kg-m^2/s^2 = Pa-m^3, you need to make sure you convert pressure units from atm to Pascals (1 atm = 101,325 Pa), and volume units from liters to m^3 (1000 L = 1m^3).  \nw = -(1.00 atm)(20.36 L - 1.20 L)(101,325 Pa/1 atm)(1 m^3/1000 L) = -1941 J\n\nThe total energy change is dE = q + w = (-350) + (-1941) = -2291 J
532405
The heat capacity of something tells you how much energy you need to add to that something in order to raise the temperature by one degree.  In this problem, we know a temperature change, and we know (or can look up) the heat capacities of the things undergoing the temperature change.  That means we can calculate how much energy was added to the system (in this case by the energy corresponding to the enthalpy change of a reaction.\n\nLet C_s be the heat capacity of the sample, and C_c be the heat capacity of the calorimeter. Then the total heat capacity of the system is simply C_s + C_c.  If the temperature change is delta-T, then the energy change is given by:\n\ndelta-E = delta-T * (C_s + C_c)\n\nFor a process carried out at constant pressure, the change in energy is equal to the change in enthalpy, so delta-E = delta-H\n\nThe specific heat of water is 4.1855 J/(gm * K) (see source), and its density is ~1 gm/cm^3.  The combined volume of solution in the calorimeter is 100 mL, and we are told to assume that the density and specific heat of this solution is the same as pure water.  The heat capacity of the solution is therefore:\n\n100 cm^3 * 1 gm/cm^3 * 4.1855 J/(gm* K)= 418.55 J/K\n\nThe total heat capacity of the system is:\n\n(418.55 + 12.1) J/K = 430.65 J/K\n\nThe temperature change was 6.1 degrees C = 6.1 K, so\n\ndelta-E = delta-H = 430.65 J/K * 6.1 K = 2627 J\n\nNow, this enthalpy change was for the reaction of 50 mL of a 1M CuSO4 solution with 50 mL of 2M KOH solution.  50 mL of 1M CuSO4 solution only contains 0.05 liters * 1 mole/liter = 0.05 moles of CuSO4.  If we want the enthalpy change PER MOLE of CuSO4 reacted, we have to divide the enthalpy we just calculated by 0.05:\n\ndelta-H (J/mole CuSO4) = (2627 J)/(0.05 mol CuSO4) = 5.254*10^4 J/(mole CuSO4)
607758
W=Qh-Ql=(1.23-0.986)*1e5\n=2.44*10^4  (...using 1st Law of Thermodynamics)\nP=W/t=2.44*10^4/60 =406.7 watt\n\nConsidering COP(coefficient of performance) as the energy efficiency rating,\nCOP=Ql/W\n=9.86/2.44 = 4.04\n\nHere, Ql stands for heat taken from source at low temperature(room), Qh for heat rejected to hot sink(outside air) and W is the work done by the unit.
347105
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram-°C.  If you want to use 4.18, that is okay.\n\nThe heat added = the specific heat of water (c) times the mass of the water (m) times the change in temperature.\n\nQ = cmT or\n\n= 4.18 joule/g-°C * 125 g * 2.44 °C\n= 4.18 * 125 * 2.44 joules (you can do the math)\n\nThis is for 1.75 g of CaCl2 which has a molecular mass of 111.0 g/mol, so the number of moles is:\n\n1.75 g / 111.0 g/mol = 0.01577 moles of CaCl2.\n\nDivide the heat change (in joules) above by 0.01577 moles to get the heat per mole.
444854
The answer should be A.\n\nAverage kinetic energy of a substance is related to the speed of the particles moving in it and the temperature of the substance.\n\nAs the particles in the a substance moves faster, the average kinetic energy of the substance increases.  So does the temperature.\n\nWhen a substance changes phase, the temperature will remain constant.  This means that the average kinetic energy is also constant during this process.  Hence, B is wrong.\n\nWhen thermal energy is being transferred out of the substance, one could understand it as taking out some particles out of the substance.  So what is changed here?  I would think that the total kinetic energy is being changed, since some particles have been taken away.  But the taking away of the particles would not affect the speed of the remaining particles.  The AVERAGE kinetic energy should remain unchanged.  Hence, C is also wrong.
442035
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.
74753
See reference. Calorimetric constant is the heat capacity of the inside of a calorimeter.
29685
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
478180
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......