Question

A 40 g super ball strikes a wall with a velocity of 10 m/s that is normal to the wall it bounces away at a velocity of 7 m/s so normal to the wall what is the boss change in momentum is the ball bounces last is. One seconds what is the force between the ball and the wall

Answer 1

... The change in the ball's momentum is

                (40 g) x (10 m/s + 7 m/s) = (40 g) x (17 m/s)

                                                        =   680 gram-m/s

Since we have to work with the impact formula in the next part,
we ought to change the grams in this answer to kilograms.

                                             680 gram-m/s  =  0.68  kg-m/s . 

... If the ball is in contact with the wall for one second, then the 
force exerted on the ball by the wall is found by remembering
that the change in momentum is equal to the impact  (force x time) .

                    0.68  kg-m/s = (force) x (1 sec)

Divide each side by  1 sec :

                     0.68 kg-m/s²  =  0.68 Newton  =  the Force

And now, if I may take the liberty of making a suggestion . . .
You need to remember the many benefits of "proofreading".
That is, before you post your question, look back and read
what you have written.  That way, you have the chance to
correct any accidental mistakes that may have snuck into it,
and make it a lot easier for others to read the question and
come up with an answer.

I have to confess, I'm not completely sure of what your question is.
I know that what I have written is correct, but I honestly can't be sure
that it answers the question you asked.