A cold pack manufacturer has decided to change the chemical in their cold pack from ammonium nitrate to ammonium chloride, a safer chemical. The original cold pack formulation used 100.0g of room temperature (27.0C) water and 35 g of ammonium nitrate.
Assume that the solution has the same density and specific heat capacity as water and that any heat lost to the calorimeter is negligible. The molar enthalpht for ammonium nitrate is 15.7 kJ/mole and for ammonium chloride 14.8 kJ/mole.
I don't see a question here.
To calculate the amount of heat released or absorbed during the dissolution of the cold pack chemicals, we can use the equation:
q = m * c * ΔT
where:
q is the heat absorbed or released (in joules)
m is the mass of the solution (in grams)
c is the specific heat capacity of the solution (in J/g·°C)
ΔT is the change in temperature (in °C)
Since the solution consists of water and the chosen chemical, we need to consider both components separately.
First, let's calculate the amount of heat released when the ammonium nitrate dissolves in the water.
1. Calculate the molar mass of ammonium nitrate (NH4NO3):
- Nitrogen (N) has a molar mass of 14.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- The molar mass of ammonium nitrate is:
(14.01 g/mol) + 4 * (1.01 g/mol) + 3 * (16.00 g/mol)
2. Determine the number of moles of ammonium nitrate used:
- Divide the mass of ammonium nitrate by its molar mass.
3. Calculate the heat released by the dissolution of ammonium nitrate:
- Use the equation q = m * c * ΔT, where m is the mass of water and ΔT is the temperature change.
Repeat the same process to calculate the heat released when the same amount of ammonium chloride is dissolved.
Finally, compare the two values to determine whether the change in chemical resulted in a different amount of heat released or absorbed.