In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 5.10 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol .

Answer : The final temperature of the solution in the calorimeter is,

Explanation :

First we have to calculate the heat produced.

where,

= enthalpy change = 82.8 kJ/mol

q = heat released = ?

m = mass of = 5.10 g

Molar mass of = 110.98 g/mol

Now put all the given values in the above formula, we get:

Now we have to calculate the final temperature of solution in the calorimeter.

where,

q = heat produced = 3.80 kJ = 3800 J

m = mass of solution = 100 + 5.10 = 105.10 g

c = specific heat capacity of water =

= initial temperature =

= final temperature = ?

Now put all the given values in the above formula, we get:

Thus, the final temperature of the solution in the calorimeter is,

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.