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Never mind.  I think I got it....

i saw someone posted a question similar to one im working on, and i was wondering where one of the numbers come from, the 273, where does that calculation come from? and is that the right way to solve for enthalpy, i thought you multiply the numbers by each other Heat = mass x T change x specific heat?

when 2.76g (0.0200 mol) g K2CO3 was mixed with 30.0 ml approximately 2M HCl, the temperature rose by 5.2 C

B) calculate the enthalpy change (^H) of this reaction per mole of potassium carbonate. Assume that the specific heat of the final mixture is 4.184 J/g C, and that its density is 1.00 g/mL
Q= 5.2 + 273 = 278.2
H=?
m= 2.76
2.76= moles of K2CO3
H= 278.2/2.76= 100.79 K J/Mole

Thank you so much for your help, but I have another question with this problem:

When I plug these in, I get:  4x^2 / .4-1.3x+x^2 = 54.3

I plug the bottom into a quadratic formula and get x = 0.375

is this the x for the original ice table?
I tried plugging "x" back into the equation to see if i reached 54.3 and did not
I am confused, what do i do with the 0.375 I get from the quadratic formula?

It has been 24 years since I took Algebra so a lot of these concepts need to be refreshed with me

calculate the pH of this aqueous solution? .600 moles of sodium acetate and .200 moles of HCl in 1.50 Liters solution

I keep getting that the pH is zero, is that right?

Consider the aqueous solution and determine the pH.

.500 moles of acetic acetic acid and .850 moles of NaOH in .800 Liters solution?

Isn't this a weak acid an strong base? what do i do next?

Thank yo so much for your help, but I have another question with this problem:

When I plug these in, I get:  4x^2 / .4-1.3x+x^2 = 54.3

I plug the bottom into a quadratic formula and get x = 0.375

is this the x for the original ice table?
I tried plugging "x" back into the equation to see if i reached 54.3 and did not
I am confused, what do i do with the 0.375 I get from the quadratic formula?

Fat molecules are non-polar and do not dissolve in polar solvents such as water. Hexane is non-polar and an excellent solvent for fat.

Yes an ICE table is needed.

   H2(g)    +    I2(g) <-->   2HI(g)
  1 mol          1 mol          2 mole

For each n mole of H2 that react, n mole of I2 also react and you will get 2n moles of HI

                 [H2]            [I2]                  [HI]
Initial          0.8             0.5                    0
Change        - n            - n                  + 2n
Equil         0.8-n          0.5-n                 2n

Plug these equil. values into Kc = 54.3 = [HI]²/[H2][I2] and solve for w.

You will need to solve a quadratic equation

Why would hexane be used as a solvent to remove fat from a fast food item?

it makes sense when you explain it like that, thanks!