what is enthalpy in layman's terms?
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The sum of the internal energy of a body and the product of its volume multiplied by the pressure.
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Would you like to qualify that question with something?\nFor example, what material can absorb heat with the smallest increase in temperature.\n\nHeat is a form of energy.  Temperature is the measure of the average kinetic energy of the molecules of a substance.\nAll materials will absorb heat (energy) and their temperature will increase.  How much their temperature increases when a given material absorbs a given amount of heat depends on its specific heat.\nThe specific heat of a material lets up know how much a substance's temperature will increase after abosorbing a given amount of energy.\n\nThe change in temperature of substance experiences can be calculates using the following,\nQ = m*c*(delta T)\nWhere Q is the heat energy, m is the mass of the object, c is the specific heat, and (delta T) is the change in temperature.\n\nA low specific heat means the material will experience a larger increase in temperature (all other things being constant) than a material with a higher specific heat.\n\nThe specific heat of all materials vary.\nAs a general rule, metals tend to have low specific heats and gasses tend to have higher specific heats (there is a reason for this, but it is beyond this answer's scope), but this is not always the case.\n\nWater for example has a pretty high specific heat at about 4.186 J/g degree C (or to simplify, 1 calorie).  By comparison, lead metal has a specific heat of only  1.28 J/g C.\nThis means that the temperature of 1 gram of water will increase by 1 degree C when it absorbs 4.186 Joules of energy.  Likewise, when water cools 1 degree C, it releases 4.186 J of energy.
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Now, this answer depends on what you mean by Heat.  Strictly speaking Heat (symbol 'Q')is the energy transferred from one part of a substance to another, or from one object to another due to a change in temperature.  To give you an example, if you heat a spoon on a flame and then insert it in a glass of water, the latter will warm up since Heat moves from the hot spoon to the cooler water.  Heat is a measure of energy and so its units are Joules (J).  The amount of Heat energy transferred is calculated by the formula Q=mcdT (mass x specific heat capacity x change in temperature)\n\nNow, if you are still studying basic Physics, you might come across definitions of Heat which state that it is the Energy driving the vibrations and movement of molecules.  In further Physics this will be called "Internal Energy", symbol 'U'.  In fact, the internal energy of a body can either be changed by giving or taking Heat (with a temperature difference) and/or by doing Work (that is why if you rub your hands together they warm up, despite being at the same temperature).\n\nTemperature is a measure of the degree of hotness of an object (not a measure of the amount of Heat in it!).  So, if an object has a temperature of 0 degree Celsius, this means that this object is as "hot" as pure melting ice.  In the same manner, a stone at a temperature of 100*C is as "hot" as steam above boiling water, and so on.  So, temperature is really a comparison.  There is an Absolute Temperature Scale devised by Kelvin in 1848, who found out that the lowest possible temperature is -273*C.  At this temperature all molecular movement is thought to stop and therefore there cannot exist a place colder than a place where there is no Internal Energy, and hence no heat!  Therefore this temperature is called Absolute Zero, and a temperature scale starting from this point, 0 K(elvin) (=-273*C), was set up. (e.g. 10*C = 10 + 273 = 283K).
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The study of heat and how it behaves.
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exothermic is the release of heat to the surroundings. endothermic is the absorption of heat from the surroundings.  Your temperature went up...think about it.  Part 2) specific heat of water is 4.18 j/gK, you have 55g substance. and temperature change was 9.7K...do the math.\nPart 3) well you have qH2O...it is the negative of deltaH!!\nPart 4) you have half as much NaOH...Half as much deltaH\nPart 5) divide delta H from part 3 by the number of moles NaOH in 2g NaOH
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Standard Enthalpy (H) is defined by the enthalpy change of formation for 1 mole of reactant.\n\nIf you use 5.3 moles of CO2 as the limiting reagent, you'll have 5.3 times the change in enthalpy.  5.3 X 117.3 = 621.69 kJ\n\nWith 34.5 g of CO2 as the limiting reagent, you need to start by calculating the number of moles of CO2.  1 carbon is 12g/mole.  2 O is 32 g/mol.  CO2 is (12+32) 44g/mol.  34.5g divided by 44 g/mol = 0.78 moles.  Therefore, you'll have less than the standard enthalpy change.  You calculate the enthalpy as you did above multiple 0.78 moles by 117.3 kJ/mol = 91.49 kJ.
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Specific heat cannot be negative. The specific heat of a substance is the amout of heat necessary to raise 1 g of a substance by 1 degree celcius in temperature.  You might be thinking about enthalpy with a negative number, because when a reaction is exothermic, the enthalpy change, or ΔH, is less than zero.  You can use the specific heat of a substance of known mass to calculate the enthalpy of a reaction in a closed system (for example, in a bomb calorimeter you use the specific heat of water, mass of the water, and the temperature change of water surrounding a submerged reaction vessel to find enthalpy of the reaction).
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A calorie is the ammount of energy needed to raise the temp. of one ml of water one degree C.  A Calorie is 1000 times that, or the energy needed to raise one L. of water one degree C.
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Not sure the level with which you are learning, so feel free to ignore the thermodynamics explanation.  If you just want to answer the question, select D.\n\nLike most test questions, there is not enough conditions specified and truly, we could choose several anwers, but only after making assumptions that less likely to be in the testers mind.  I will assume a high school or intro level college course because above that this should be challenged as too vague.\n\nUse logic to realize that answer choice C is not true. Evaluating the others we realize that A means only heat is given off when bonds form and B means heat is absorbed or added during the formation of bonds.  D means conditions exist in which either situation is possible and E means that final product is in a state that has a higher value for entropy.  This only holds true when both Temp and pressure of a system is held constant, but we will eliminate later as well.\n\nBasically a bond can occur when the system goes from a high energy state to a lower energy state.  Formation of bonds is essentially a chemical reaction.  The Gibbs Free Energy equation:\n\n (delta)G=(delta)H-T(delta)S\n\nT=temp of system in Kelvin\n(delta)H is the heat energy change\n(delta)S is the change in entropy (disorder)\n\ndefines whether a reaction will occur spontaneously or not.  (deltaG) must be negative for this to be true. So we will assume \nusing this equation and recognizing that (delta)H tells us exo or endothermic if it is positive or negative , examination of the possibilities of (delta)H and (delta)S,\n\nIf (delta)H is negative and (delta)S is positive the equation will result in a negative value for (delta)G which means a favorible condition exists for bonding to occur.  Therefore, the condition of (delta)H negative (energy leaves as heat - exothermic) is true.  We can thus eliminate answer choice B.\n\nIf the opposite is true and (delta)H is positive and (delta)S is negative, the result is (delta)G that is positive and therefore not spontaneous.  no reaction or bonding occurs.\n\nIf both (delta)H is positive and (delta)S is positive, only a large enough (delta)S will overcome the positive (delta)H in the equation.  So if both are positive and T(delta)S is greater than (delta)H the bonding will occur.  Therefore, when (delta)H is positive (endothermic) the reaction may occur as well.  We can thus eliminate choice A.\n\nWe have shown that C is a true statement because conditions can be met in which the reaction is both exo or endothermic.  However, please note that in both conditions (delta)S is also positive, indicating an increase in entropy.  For the most part any system will always trend toward an insrease in entropy.  \n\nThe question then arises, what happens when (delta)S is negative (decreased entropy) and (delta)H is negative.  For most situations T(delta)S will be large enough to create a positive (delta)G from our equation.  In very rare cases will (delta)H - T(delta)S result in a negative answer.  However, the equation shows this as a possibility when (delta)G= large negative (delta)H - T x small negative (delta)S => large negative # minus a smaller negative => larger negative + smaller # = negative.  (Hope you follow my train of thought).  Therefore in the right situation - (delta) can be negative and bonding/reaction can occur when entropy decreases.  Eliminate E and C is the correct answer.\n\nCareful what you ask for.  Wikipedia site looks pretty informative but very detailed and you have to use a few different pages.
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Standard enthalpy of formation 227 KJ/mol\nStandard enthalpy of combustion -1300 KJ/mol\nStandard entropy 201 J/K/mol
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Calorimeter Info center\n\nhttp://www.scipeeps.com/Sci-Chemistry_Topics_C_-_Cl/Calorimeter.html
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heat is a form of energy.
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OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
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i already answered parts a and b in the last question like this.\n\nc) enthalpy = mass of water x specific heat of water x change in temp\n\nE=0.1*4.18*3.5\nE=1.463KJ\n\nmultiply by two to get answer per mole = 2.926KJ/mol\n\nd) main source of error is the glass beaker since it is leak out the heat produced by the reaction, the reaction would be better carried out in a calorimeter or in a polystyrene beaker with a lid.\n\ne)  the temperature change would be the same as 1mol of KOH reacts with 1mol of HCl therefore with the extra HCl the KOH will act as a limiting reagent and u will get the same reaction as before.
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2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
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Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
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HEAT ENERGY=mQ(t2-t1)\n\nM-MASS OF SUBSTANCE\nQ-SPECIFIC HEAT CAPACITY\n(t2-t1)-CHANGE IN TEMP.