What formula do I use to figure out the final temperature of a mixture?
819865
Conservation of energy.\n\nCalculate the amount of heat stored in the 28ºC water using 0ºC as the starting point.  This energy must first melt the ice, for which you'll need to use the heat of fusion of water.  The remainder of the energy will warm up the entire amount of water.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
76548
mc(delta T) = -mc(delta T)\nTf = the equilibrium temperature (final temperature) and is the same for both metals\n\n(2.6kg)(.22kcal/kgC)(Tf-13C) = -(10.2kg)(.093kcal/kgC)(Tf-93C)\n\n.572(Tf)-7.436 = -.9486(Tf)+88.2198\n1.5206(Tf) = 95.6558\n\nTf = 62.91 C
864388
some formulas that can help :\nhttp://www.eskridgeinc.com/support/formulas.html
278691
Use the combined law of gases formula P1*v1/T1=P2*v2/T2.  Where P=pressure, v=volume, and T=temperature.  If the problem doens't have either the temperatures or the pressures assume them to be 1.  I.E. The answer to problem number 7 is:  8 liters.
252754
ok ill try to explain cuz the guy before didnt. its all basically the same kinda questions.  using pv=nrt, 2 things stay constant  and 2 things change. and you put everything thatz constant on the right side, and then everything thatz changing on the left side. and you get a proportion relation, something like  p1v1 = p2v2  (for number 1 and 2), or p1/t1 = p2/t2. \nand the units are not important at all. the only units that are important is you have to have temperature in kelvin. \n\n1. so pv=nrt, and nrt stays constant in this. if it doesnt say anything about the number of mols changing or the temperature changing, assume its constant. \nso then p1v1 = p2v2. \n(99 kPa)(300mL) = (188kPa)v2\nv2 = (300mL) (99kPa) / (188kPa)\nv2 = 157.9787mL.\n\n2. same thing. p1v1 = p2v2\n(0.988atm)(1L) = p2 (2L)\np2 = (0.988atm) * (1L)/(2L)\np2 = 0.494atm.\n\n6.  so from pv = nrt. u gotta get all the constants on the right side. in this case, the constants are P and n. so divide both sides by p and t. \npv/pt = nrt / pt\nv/t = nr/p.     so if v changes, then t has to change accordingly to make sure that v/t is equal to nr/p at ALL times.\nso v1/t1 = v2/t2. \n(0.67L) / (89+273K) = (1.12L) / t2  \n(89+273K) * (1.12L) / (0.67L) = t2\nt2 = 605.134K\nt2 = 332 degrees celsius.
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.
141221
The original water temp is 25C, as was given in the problem... I'm assuming you meant to ask what was the initial temp of the water that was added, which isn't too hard. Here goes.\nQ(heat)= m(mass) x c(specific heat) x delta T(change in temp)\nso Q = (150g)(4.something J/gC-look it up)(25C-14C)\nBecause of conservation of energy and assuming ideal insulation, the heat LOST by the 150g of water during the mixing is equal to the heat GAINED by the 200g of water. So, once you have Q, rearrange the equation to look like [delta T] = Q/mc = Q/(200g)(c)\nThat will give you the temperature difference between 14C and the 200g of water's initial temp. Delta T should come out as being positive, since the temp is going up. \nWhen you have more complex mixtures of things with different specific heats and add processes and changes, the math becomes a lot more of a headache but the same basic principles apply just the same.
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
773838
The formula would be 2HCL+ CaCO3 ---> CaCl2 + H2O + CO2. The standard enthalpies of formation are -167.2, -1207, -796, -285.8, and -393.5 respectively. When you use the formula "Sum of products minus sum of reactants, you come up with +66.1 (don't forget to multiply the HCl formation by 2 for 2 moles). I am assuming that the HCl you are talking about is HCl (aq) and not HCl (g). If it is a gas, then the standard enthalpy of formation is -92.31, so you would come up with a different number (-83.68)\n\nIf it is aqueous HCl, you come up with a positive 66.1 kJ/mole, which means that heat was absorbed by the reaction (endothermic). That probably means your cylinder would feel cooler, since heat is being taken in by the reaciton.\n\nIf you mean gaseous HCl, you come up with a negative 83.68 kJ/mol, which means that heat was given off by the reaction (exothermic). That probably means yoru cylinder woudl feel warmer, since heat is being given off by the reaciton and taken in by the cylinder.
883220
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......
684353
P1V1T2=P2V2T1 \n\nRemember to express  temperature in Kelvin (with pressure in Pascals and volume in m^3)
856026
Use the gas law\nPV=nRT
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
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.