Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

a 50 ml sample of a 1.00 M solution of CuSO4 is mixed with 50 ml of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 degrees Celsius before mixing and 26.3 degrees celsius after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data calculate delta H for the process. Assume the specific heat and density of the solution after mixing are the same as those of pure water.

Sagot :

Answer:

Explanation

Solution:- Total volume of the solution = 5.0 mL + 50.0 mL = 100.0 mL

Density of solution is same as of water that is 1.0 g per mL. So, the mass of solution, m =  

m = 100.0 g

change in temperature,  = 26.3 - 20.2 = 6.1 degree C

specific heat, s of water is 4.184 J per g per degree C.

where q is the heat energy. let's plug in the values in the equation:

q = 2552.24 J

energy of calorimeter = change in temperature*heat capacity of calorimeter

energy of calorimeter =    

= 73.81 J

Total heat = 2552.24 J + 73.81 J = 2626.05 J

Let's convert it to kJ and for this we divide by 1000 since, 1000 J = 1 kJ

= 2.626 kJ

0.05 moles of copper sulfate are used so the  of the reaction =  

=  

Since, there is an increase in temperature, the heat is released and so the sign of  will be negative.

Hence,  = 52.52 jk