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Mass of beaker plus solid Mass of water,mno 1.229 4,298-g 41-949 Mass of solid,

ID: 593958 • Letter: M

Question

Mass of beaker plus solid Mass of water,mno 1.229 4,298-g 41-949 Mass of solid, m Initial temperature Final temperature a1,0 for the reaction (Eq. 1 ) (SH. = 4.18 J/g°C) H for the reaction (Eq.5) The quantity you have just calculated is approximately" equal to the heat of solution of your sample. Calculate OC - joules joules the heat of solution per gram of solid sample. The solution reaction is endothermic exothermic. (Underline the correct answer.) Give your reasoning. Solid unknown # s ndol g/mol oal Formula of compound used (if furnished) Heat of solution per mole of compound C. Heat of Neutralization Molar mass kJ/mol °C Initial temperature of HCI solution Initial temperature of NaOH solution Final temperature of neutralized mixture Change in temperature. Ar (use the average of the initial temperatures of oC 3/ °C HCI and NaOH) u,o (assume 50.0 mL of solution and use average density of 1.02 g/mL.) Total H for the neutralization reaction AH per mole of H and OH ions reacting

Explanation / Answer

Heat of solution

Mass of water = 48.298 g

mass of solid = 4.999 g

Total mass = 53.297 g

specific heat of water = 4.184 J/g.oC

Initial temperature = 20 oC

Final temperature (assumption since data is not given above) = 16 oC

dT = 16 - 20 = -4 oC

q(water) = mCpdT

              = 53.297 x 4.184 x -4 = -892 J

The final temperature is lower than initial temperature, so reaction is endothermic in nature

dH(soln) = -q(water) = 892 J

dH(soln)/g solid = 892/4.999 = 178.435 J/g

Solid identity = KNO3

dH(soln)/mole solid = 892 J x 101 g/mol/4.999 g x 1000 = 18.022 kJ/mol

[pl. note final temperature is assumed here, you must feed the correct value to get the answer]

Heat of neutralization : HCl + NaOH

Mass of solution = 50 ml x 1.02 g/mol = 51 g

initial temperature = (20 + 19)/2 = 19.5 oC

final temperature = 31 oC

dT = 31 - 19.5 = 11.5 oC

q(water) = mCpdT

              = 51 x 4.184 x 11.5 = 2453.916 J

dH for neutralization = -q(water) = -2453.916 J

Since molarity of solution is missing, we cannot find dH/mole H+ or OH-