Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Discover detailed answers to your questions from a wide network of experts on our comprehensive 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