Asked by Wendy
I posted this question earlier but the person that answered did not include how to find the moles of the problem
When 80.0 mL of HCl is added to 80.0 mL of NH3OH, the temperature increases 2.5 celsius. Assuming the final solution has the same specific heat capacity as liquid water, the heat produced in the reaction between two solutions is:
I don't know how to find the moles like i said before.
When 80.0 mL of HCl is added to 80.0 mL of NH3OH, the temperature increases 2.5 celsius. Assuming the final solution has the same specific heat capacity as liquid water, the heat produced in the reaction between two solutions is:
I don't know how to find the moles like i said before.
Answers
Answered by
Wendy
Sorry I meant NH4OH, not NH3OH
Answered by
DrBob222
I answered this before and I told you how to work the problem. You don't need mols.
The heat produced by the reaction is
q = mass x specific heat x delta T.
You know mass (160g), you know specific heat, and you know delta T. You don't need mols if you use mass in grams and specific heat in J/g*C (4.184). If you persist in mols and you have specific heat in J/mol and you don't know how to change specific heat from J/g to J/mol, then you have 160 g water and that is 160 g/molar mass H2O = mols H2O. But I wouldn't go that route, that's changing more units than you need to change.
The heat produced by the reaction is
q = mass x specific heat x delta T.
You know mass (160g), you know specific heat, and you know delta T. You don't need mols if you use mass in grams and specific heat in J/g*C (4.184). If you persist in mols and you have specific heat in J/mol and you don't know how to change specific heat from J/g to J/mol, then you have 160 g water and that is 160 g/molar mass H2O = mols H2O. But I wouldn't go that route, that's changing more units than you need to change.
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