calorimetry: disolving salts, please help?
959876
a) the hydration energy is greater because the temp decreases. When you go from a low potential energy substance to a high potential energy substance the temp of solution will decrease.\n\nb) the solubilty will increase because energy is on the reactant side of the equation. (the reaction is endothermic and requires energy to occur)
945741
Due to the fact that this salt is found as a hydrated salt (first Na2SO4.10H20 and then as Na2SO4.7H2O, so, with the increase of temperature it looses moisture (water)). When the Na2SO4.10H2O is heated the solution of this compound decreases because a solid phase and a liquid phase are formed after the initial heating.
986555
Here you have the heat of solution of sodium hydroxide. LOL, it may help you with your work:\n\nhttp://chemlab.byu.edu/worksheets/0cb1f0c84fee4d6cda5d5f99832e9513.pdf\n\nand you can find the heat of solution for ammonium chloride here:\n\nhttp://www.brynmawr.edu/Acads/Chem/Chem103/swhite/exam2answersfall99.html\n\nThe heat of solution of NH4Cl is + 14.7 kJ/mol.
811655
If by 'undissolved' you mean why are some salts insoluble in water, then I think we are in business.\n\nIn order for a salt to be soluble it has to be energetically favorable for the ions to separate. This requires a negative gibbs free energy. G = H - TS So you have to find the H and S values for the solid to aqueous solution and see if G will be negative for a given temperature.\n\nA method I recently learned to make insolube salts more soluble is to add other ions to the water...but you have to be careful to not add a common ion...or Le Chantliear will come into play.
164495
this experiment is called calorimetry
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
838309
b a solution, the salt will dissolve
812576
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
1051035
1-. NaCl + H2O Sodium Chloride ( table salt )\n2-. NaSO3 +2 H2O Sodium Sulfate ( or sulfite I always forget which is the 3 and which is the 4 )\n3-. NaC2H3O2 + H2O ( I'm guessing.  You will want to check your periodic table for the charges on this one )\n4-.NH4Cl + H2O.  Ammonium Chloride\n5-.NSO2 + 3H2O\n6-. not sure\n\nI presume you are working salt solutions.  Ammonia was never my strong suit. I'm positive about the first two.
731293
sodium & chlorine to make NaCl, table salt
288276
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
389448
Na+ and OH- you are dissolving a salt\n\nby the way, i am getting a phd in chem, so trust me. also, there are tons of answers in there that are totally bullshit.
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
507893
1 calorie heats 1g of water by 1 deg C.\n630cal heats 630g of water 1 deg C\n630cal heats (630/15)g water by 15deg
362368
Given:\nm <mass of substance> = 12.0g or 12/1000 kg\nc <specific heat capacity of water> = 4180J per kg per degree celsius\nt <change in temperature> = 20 degrees to 100 degrees i.e. 80 degree celsius\n\nFind:\nHeat req.\n\nSolution:\nHeat required or Q\nQ = m * c * t\n  = 12/1000 * 4180 * 80\n  = 12/10 * 418 * 8 (cancelling 0s of 4180&80)\n  = 12/5 * 209 * 8 (div. 418&10 by 2)\n  = 12/5 * 1672 (mult. 207&8)\n  = 20064/5 (mult. 12&1672)\n  = 4012.8\n\nAnswer = 4012.8 Joules or 4.0128 KJ
1064976
Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
170755
Specific heat 0.13 J/gK