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a laboratory technician adds 43.1 mL of concentrated, 11.6 mol/L hydrochloric acid to water to form 500 ml of dilute solution. the temperature of the solution changes from 19.2C to 21.8C. calculate the molar enthalpy of the equation.

so this is what is given

t=temperature (19.2-21.8= -2.6)

c=specific heat capacity. for water is 4.18 but i dont know if im suppose water for this

m=mass i'm a bit lost in this one but i used 43.1ml

/\H=mc/\t/n

this is what i tried

mHCl=43.1g t= -2.6 MHCl= 36.46 g/mol

43.1/36.46=1.18n mol

(43.1*4.18*-2.6)/ 1.182

=150.2

and i think the answer i got is wrong

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# Calorimetry question

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