enthalpy of vaporization?
604334
since vapourisation is a change in state, it can also be called the heat of vapourisation, q.\n\nΔS = q/T\n\nΔS = (30x1000)/(273.15+76.8)\n\nΔS = 85.73 J/mol/K\n\nThat answer is only correct to a certain degree, since the temprature changes during the process, but it does.  otherwise you have to do a lot of maths.
871634
stndrd enthalpy of formation for NaF (s) is -569.0 kJ/mol\n\nhttp://chemistry.about.com/library/weekly/blheatform.htm
85318
Heat is equal to the change in enthalpy, dH, which is equal to the enthalpy of the products minus the enthalpy of the reactants.\n\ndH=H(products)-H(reactants)\n\n2B + 3F(2) --> 2BF(3)\n\nGet the heat of formation values for each specimen from a chemical thermodynamics table. I don't have a table in front of me so I'm just going to make up values so you'll know how to solve the problem. Let H=100 for B, H=150 for F(2), and H=400 for BF(3).\n\ndH= 2(400)-[2(100)+3(150)]\ndH=800-650=150 kj/mol\n\nI do know that elements alone have a heat of formation of 0, but I don't remember if the element has to be in a certain state or not.
983222
i dont really feel like thinking about the question but this should help\nuse the fact that the boiling point is when the vapor pressure equals the atmospheric pressure \nso at 139.6C the vapor pressure is 1 atm, perhaps you can just set up a ratio, but since that delta h is there u need to use something that involves that \nsrry that im feeling lazy
211379
Vapor Pressure\n\n\n\nOne of the physical properties that we can measure for liquids and solids is vapor pressure. Vapor pressure is defined as the pressure of the vapor of a liquid or solid in equilibrium with the liquid or solid state. Generally liquids have higher vapor pressures than solids. \n\nYou can look at vapor pressure from a molecular view. Let's say that we have a glass of liquid water. The only molecules of water that have the highest probability of escaping into the vapor phase are the surface molecules. Molecules of water below the surface have forces between the molecules called Intermolecular Forces that are stronger then these forces exert on surface molecules. This is simply because there are more molecules surrounding the interior water molecules than what surround the surface molecules. As a result, water molecules can only escape at the surface. The physical process by which vapor pressure occurs is referred to as evaporation. There are several things that could influence the rate at which liquid molecules become vapor. \n\n\nPolar Nature of the molecules \nIf the molecules are non-polar then there will be weaker intermolecular forces keeping them in the liquid state. Non-polar substances like hydrocarbons have weaker forces between them and therefore their vapor pressures are characteristically higher than polar substances. Polar substances have dipolar molecules with positive and negative ends to the molecules. The intermolecular forces are much stronger and the liquid molecules are kept in the liquid state easier. Polar substances tend to have lower vapor pressures at a given temperature. The rate of evaporation for non-polar molecules would be generally greater than polar molecules in comparison. \n\n\nSurface area \nSurface area would play apart since evaporation is a "surface phenomena" as was mentioned above. It stands to reason that if we increase the surface area of a liquid that we will increase the number of surface molecules and therefore more molecules will be able to escape into the vapor state. This is why when we want to evaporate something we place it on a watch glass or in a beaker and not in a test tube where the surface area is much more restricted. Increasing the surface area will increase the rate at which something will evaporate. Unlike the polar nature of the molecules, changing the surface area does not affect the vapor pressure itself. \n\n\nTemperature \nTemperature increases increase the available Kinetic Energy present. The more energy that the surface molecules can absorb the faster they will be able to overcome the intermolecular forces acting upon them and the sooner they will become vapor. Increasing the temperature increases the rate at which something will evaporate. Like the polar nature of molecules it also affects the vapor pressure. Increasing the temperature of a liquid increases the vapor pressure logarithmically. \n\n\nThe concentration of vapor molecules above the liquid \nIncreasing the number of vapor molecules above the liquid will leave less room for further vapor molecules. This has a depressing effect on the rate at which the liquid evaporates. Reducing the concentration of vapor molecules at the surface will increase the rate. That is why we turn on a ventilation fan under the vent hood of a lab when we wish to evaporate something quickly. In addition, clothes hanging on a clothes line dry faster on a windy day then on a still day. \n\n\nThe Vapor Pressure of a liquid at a given temperature is a characteristic property of that liquid. It can be measured by using a pressure instrument known as a manometer. Vapor pressure of a liquid is intimately connected to another physical property known as boiling point, but that will be another topic for discussion. \n\nVapor Pressures are influenced by Temperature logarithmically and this relationship is defined with the Clausius Clapyron Equation: \n\nLog P2 / P1 = Delta H vaporization [ 1 / T1 - 1/T2
968956
It appears what you really need is the heat of vaporization.  Cp lets you calculate the amount of heat needed to raise or lower the temperature of a certain mass of liquid, but that's insignificant compared to the amount of heat gained or lost with a liquid-vapor phase change.
240367
Endothermic.   It takes (heat) energy to vaporize a liquid.\n\nThink of evaporation from the skin.  It feels colder when you're wet.   It takes heat energy away (endothermic) from your skin when the water evaporates.
896092
The vapor pressure of a thermodynamic system, at a given temperature, wherein the vapor of a substance is in equilibrium with a plane surface of that substance's pure liquid or solid phase.
928379
(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
464223
'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
386299
Enthalpy is a quantifiable state function, and the total enthalpy of a system cannot be measured directly; the enthalpy change of a system is measured instead. Enthalpy is a thermodynamic potential, and is useful particularly for nearly-constant pressure processes, where any energy input to the system must go into internal energy or the mechanical work of expanding the system. The change in enthalpy of a system is nearly singularly meaningful for systems at constant pressure, when the change in enthalpy is equivalent to heat. Otherwise, there is no easily-interpreted meaning for this state function. as heat of the reaction for reactions performed at constant pressure. For a simple system, with a constant number of particles, the difference in enthalpy is the maximum amount of thermal energy derivable from a thermodynamic process in which the pressure is held constant.
1019587
condensate depression: the amount of sub-cooling required to remove the latent heat of vaporization at a given vacuum.
423892
I'm sure that's what you're supposed to determine in the lab you're working on. Good luck.\n\nI've been able to find a value for the heat capacity of n-hexadecanol (cetyl alcohol) in my copy of _Lange's Handbook of Chemistry, 13th Edition_.  When you've come up with your value from the experiment, post it here and I'll check to see if you're close.
1165181
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it
953068
Evaporation is one of the two forms of vaporization. It is the process whereby atoms or molecules in a liquid state (or solid state if the substance sublimes) gain sufficient energy to enter the gaseous state. It is the opposite process of condensation.
937733
Enthalpy change of formation of 1 mole of C2H2 is the energy change when 1 mole of it is formed from its elements (2C(s) and H2(g)) in their standard states.\n\n2C(s) + H2(g) ---> C2H2(g)\n\nenthalpy of formation of C2H2= \nenergy absorbed to break H-H bond \n+ \nenergy released to form 1 C=C bond and 4 C-H bonds \n= 610 + 4(-410)\n= -1030 KJ/mol\n(If you are taking A level Chemistry, refer to your data booklet for these values under the section of bond energies)\n\nHope this helps.
159817
temperature of fire was 900°F to 1200°F\n\nMost volatiles ignite at temperatures ranging from 1000º - 1400º F.\n\ncombusted in the fourth and final state at temperatures in excess of 935°F.