Question

How many joules are needed to heat 8.50 grams of ice from -10.0 degrees to 25.0 degrees

I came up with 2124 joules

Answer 1

I believe I worked that problem yesterday. Perhaps not for you.
heat needed to move T of ice from -10 to 0
q1 = mass x specific heat ice x delta T

heat needed to melt the ice.
q2 = mass x heat of fusion.

heat needed to move T from 0 to 25.
q3 = mass x specific heat water x delta T.

Total heat required is q1+q2+q3.
Post your work if you get stuck.

Answer 2

For the specific heat of ice I looked it up and found it to be 2.05 and the heat fusion to be 334

q1 = 8.5 x 2.05 x 10 = 174.25

q2 = 8.5 x 334 = 2839

q2 = 8.5 x 4.184 x -25

and that gave me a total of 2124 joules...also for my 2nd question from yesterday...

Caculate the hydronium ion concentration and the hydroxide ion concentration in blood, the pH of which is 7.3 (slightly alkaline)

answers:
[hydronium ion]= 5.0 x 10^-8
[hydroxide ion]= 2 x 10^-7

Answer 3

q1 is ok.
q2 is ok.
q3 (which you typed in as a second q2) is not correct because you made it a negative number. Since heat is being added it must be a positive number. You should have +889.1 J and the total should have been 3902 J but check my arithmetic.
If you don't remember to make the sign + or - depending upon heat being added or withdrawn, you may want to write the formulas as
mass x specific heat x (Tf-Ti) where Tf is final T and Ti is initial T.
For 0 to 25 that will be
8.5 x 4.184 x (25-0) = a + number
It works for the -10 to 0 also.
8.5 x 2.05 x [0-(-10)] = 8.5 x 2.05 x 10 = a + number.
I don't remember seeing the blood problem.