When solutions of silver nitrate and sodium chloride are mixed, a precipitation reaction occurs. what mass of precipitate can be produced from 1.14 l of a 0.269 m solution of silver nitrate reacting with excess sodium chloride?

Explanation:

The given precipitation reaction will be as follows.

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity =

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

                    Molarity =

                     0.269 M =

                    no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

               No. of moles =

                   0.307 mol =

                           mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.

c. orbital shape i hope this is correct

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.