A calorimeter contains 35.0mL of water at 15.0 degrees celcius . When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)--->X(aq) and the temperature of the solution increases to 29.0 degrees celcius . Calculate the enthalpy change,(deltaH), for this reaction per mole of X. Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 (J/g(degree celcius)) and 1.00g/mL ] and that no heat is lost to the calorimeter itself, nor to the surroundings.

Prob -1  
A calorimeter contains 35.0mL of water at 15.0 degrees celcius . When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l)--->X(aq)
and the temperature of the solution increases to 29.0 degrees celcius .
Calculate the enthalpy change,(deltaH), for this reaction per mole of X.
Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 (J/g(degree celcius)) and 1.00g/mL ] and that no heat is lost to the calorimeter itself, nor to the surroundings.


In this problem heat is consumed in increasing the temperature of water


heat  Q = M*C* delta T

M = mass of water = 35.0 ml * 1 gram / 1 ml  =  35 gram   (density of water is 1gram per cm ^ 3 or 1.00g/mL  )
 
delta T = final temperature - initial  temperature
          =  29 - 15  =  14 `C
 
specific heat capacity C = 4.18 (J/g(degree celcius)
 put these value in equation we get heat  Q = M*C* delta T
Heat Q = 35 * 4.18*14   
           = 2048.2j


When 1.20g of X (a substance with a molar mass of 51.0g/mol ) is added
number of moles of substance X =  Mass / molar mass 
                                               = 1.20/51.0   
                                               = 0.0235 mol

Molar heat capacity = heat required / number of moles
                              = 2048.2 j / 0.035 mol
                              = 87048.5 jules                          (1000 j = 1 kj )
convert in kilojoules
                             = 87048.5/ 1000 kj
                             =87.05 kj                            

         Answer = 87.05 kj

Problem - 2  This one problem has me confused
I need 60.0 in scientific notation
is it 6.0x10^1 ??

your answer is correct 

in scientific notation we always place decimal point after one digit from the left

example

60000000   = 6.0 x10^7
0.00006 = 6.0 x10^ -5

and  60.0  =  6.0x10^1   =  6.0x10