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.
Sorry I meant NH4OH, not NH3OH
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.
To find the moles in this problem, we need to use the concentration and volume of the solutions.
First, we'll calculate the moles of HCl.
Since we know the volume of the HCl solution is 80.0 mL, we need to convert this to liters. There are 1000 mL in 1 L, so the volume of the HCl solution is 80.0 mL / 1000 mL/L = 0.080 L.
Next, we need to determine the concentration of the HCl solution. Assuming it is given in molarity (M), let's say it is 2.0 M.
Now, we can calculate the moles of HCl using the formula:
Moles of HCl = concentration (M) × volume (L)
Moles of HCl = 2.0 M × 0.080 L = 0.16 moles of HCl
Similarly, we can find the moles of NH3OH in the same way.
The moles of NH3OH = concentration (M) × volume (L)
Since the volume of the NH3OH solution is also 80.0 mL or 0.080 L, we need to know its concentration. However, the problem statement did not provide any information about the concentration of NH3OH. Without that information, we cannot calculate the moles of NH3OH.
Please provide the concentration of NH3OH, if available, for a more accurate calculation of moles.