The specific heat of water is at 4.184J/g * degree C. How much heat is evolved, or absorbed, when the temperature of 9.00 mol of liquid water cools from 38 degrees C to 28 degrees C?

How do I start this problem

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  1. q = mass x specific heat x delta T.
    You will need to change the mols of water to grams.

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  2. wouldnt it be 4.184J/g * degrees C

    so 4.184 J/ 162 g H2O * 10 degrees C = ?

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  3. mols = grams x molar mass
    9.00 mols H2O x (18.015 g/mol) = ?? g H2O
    q = g H2O x 4.184 J/g*C x delta T
    and delta T = (Tfinal - Tinitial) = (38-28) = -10.
    The answer will be a negative number which means that heat is evolved (and not absorbed).

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  4. -6783.72 J

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  5. ok. Actually, if we round the numbers, the last 2 should be a 3. However, if you are concerned about the number of significant figures, the 9.00 limits them to three; therefore, the -6383.73 J would be written as -6.38 x 10^3 Joules.

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