I have all of the measurement filled in. I just need help with the calculations.

Question

I have all of the measurement filled in. I just need help with the calculations. Im not sure where to put the numbers in the formulas

.33133 33 33.33.23 parately) DATA calculations sep Table Heat Capacity of Calorimeter (show 1: Temp of calorimeter and cold water before mixing Temp of hot water before mixing Temp of water after mixing 10'C water water Heat gained by cold water, qe Heat lost by hot water q er: Heat gained by calorimeter, qeu: Heat capacity of calorimeter, car Table 2: Heat of Neutralization (HCI as acid) (show calculations separately) Temp of calorimeter and NaoH before mixing (after mixing) Table 3: Heat of Neutralization (HcaH,o as acid (show calculations separately) Temp of calorimeter and NaoH before mixing AT (after mixing)

Explanation / Answer

1.we know that heat q=mCpT

Cp for water =4.184 J/gm C

Let 1 mol of water ,so mass= mol*molecular weight=1*18=18

Heat gained by cold water qgain = mCp(Tafter mixing-Tbefore mixing)=18 *4.184* (31-21) = 753.12 J

Heat lost by hot water qlost = mCp(Tbefore mixing-Tafter mixing)=18 *4.184* (43-31) = 903.744 J

qcal+qhot+qcold=0

qcal-903.744+753.12=0

(here hot water is lost the heat thats why put the negative sign )

qcal = 150.624 J

specific heat Ccal = qcal /T=150.624/10=15.0624 J/ C

2. Heat of neutralization of HCl

qcal,HCl =Ccal * T=15.06*(36-22)= 210.8736 J

qnut=-qcal,HCl = -210.8736

For CH3COOH

qcal,CH3COOH =Ccal * T=15.06*(33-22)= 165.66 J

qnut=-qcal,CH3COOH = -165.66 J

Reactions

HCl+NaOH= NaCl+H2O

CH3COOH+NaOH= CH3COONa+H2O

Assume it is 1 mol (volume is not given)

HHCl,rxn= q/mol of water = 210.8736 J

HCH3COOH,rxn= q/mol of water= 165.66 J

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