q= m x s x (delta)T\nHeat of fusion for (h20) ice is 6.01 kj/mol
119760
ρwater ~ 1g/ml\n\nq = 1000g x 4.18J/ºC.g x (37ºC-0ºC) = 155.660kJ (1.6x10^5J to correct Sig Figs)\n\nsecond part. The amount of heat required in part 1 will melt 155.660kJ / 6.01kJ/mole = 25.9 moles of ice or 468g
20984
The question asked for the MAXIMUM amount of ice, so we have to assume that the ice is at the freezing/melting temperature. Any lower temperature and it won't be the maximum amount of ice because some of the heat is wasted bringing the temperature of the ice up to the freezing/melting point.\n\n\n3500 J/s * 60s = 210,000J\n\nWater's heat of fusion = 334 J/g\n\nTotal ice which can be melted = 210,000J / 334J/g = 628.74g
361169
steam to water \n15.2 x 2.03 KJ x (158-100) = 1789.648\n15.2 x 40.6 = 617.12\n\nwater to ice\n15.2 x 4.18 x 100C = 6353.6\n15.2 x  6.02 = 19.5\n\nice at 0degree C to -50\n15.2 x 2.06 x 50 = 1565.6\n\ntotal quantity of heat\n1789.648+617.12+6353.6+19.5+1565.6 = 10345.468\n\n\ni hope it is right.. nt vry sure..
520975
Delta E = q + w. 107.6 = -305.9 + w. So, w = 413.5 kj.
410473
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
260006
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J
381050
Use the equation:\nDelataH of the reaction = DeltaH(products) - DeltaH(reactants)\n= -242 - (-191)\n= -51kJ/mol
354175
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
512036
(Triangle) or (Delta) H for this process will equal:\n\n60g * Cp of water (4.1855 J/(C*g) * (Delta T)\n\n60*4.1855*14* = 3515.82 J\n\nThe change in enthalpy or (triangle H) is 3515.82J
398824
1) Raise the temperature from -20°C to 0°C:\nMass of ice * specific heat of ice * temperature rise\n10.0g * 2.1 J/g K * 20° = 420 J\n2) Melt the ice:\nMass of ice * Heat of Fusion\n10g * 333J/g = 3330 J\n3) Raise the temperature from 0°C to 100°C:\nMass of water * specific heat of water * temperature rise\n10.0g * 4.2 J/g K * 100° = 4200 J\n4) Evaporate the water:\nMass of water * Heat of Vaporization\n10.0g * 2260J/g = 22600 J\n5) Raise the temperature from 100°C to 120°C:\nMass of steam * specific heat of steam * temperature rise\n10.0g * 2.0 J/g K * 20° = 400 J\n\n6) Addition: 420 + 3330 + 4200 + 22600 + 400 = 30950 J
188555
U have not mentioned the gas gravity to determone deviation factor to have an exact answer...generally..taking z=1\npsia ( absolute value) will also be used \nPsc*Vsc/Tsc=PV/T\n\nso: Vsc/V=PTsc/PscT=53.1/T \n\nThis is the simplest correct answer.... make is easier but worse that the process is isothermal so\n\nVsc/V=102 --> 1cu ft at TVD= 102 SCF\n\nThe most exact solution can be provided if u want..with taking temperatire and deviation,composition in account
122259
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
665249
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
28188
Convert km^3 to cm^3:\n5 km3 * 1x10^15 cm3/km3 = 5x10^15 cm3\n\n\nConvert cm^3 to g:\n5x10^15 cm3 * 0.86 g/cm3 = 4.3x10^15 g\n\n\nWater's heat of fusion = 334 J/g\n\n\nCalculate energy:\n4.3x10^15 g * 334 J/g  = 1.4362x10^18 J\n\n= 1,436,200,000,000 MJ
291413
The latent heat of fusion -- once you cool water to 0°C, you have to remove more heat to actually crystallize it.\n\nIce at 0°C requires the addition of heat to melt into water at 0°C.  That heat comes from whatever you are cooling.\n\nThe latent heat of fusion for ice is 333.6 joules/gram
277356
E = h*c/wavelength  h = 6.634x10^-34 J*s
582913
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude