A coffee-cup calorimeter initially contains 125 g water at 24.2 degrees celsius. Potassium bromide (10.5 g), also at 24.2 degree celsius, is added to the water, and after the KBr dissolves, the final temperature is 21.1 degrees celsius. Calculate the enthalpy change for dieseling the salt in J/g and kJ/mol. Assume that the specific heat capacity of the solution is 4.18 J/Cg and that no heat is transferred to the surrounds or to the calorimeter.

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To calculate the enthalpy change for dissolving the salt (KBr), we need to use the equation:

ΔH = q / m

Where:
ΔH is the enthalpy change (in J/g or kJ/mol)
q is the heat absorbed or released (in J)
m is the mass of the substance (in g or mol)

First, let's calculate the heat absorbed or released (q):

q = m_solution * C_solution * ΔT

Where:
m_solution is the mass of the solution
C_solution is the specific heat capacity of the solution
ΔT is the change in temperature

Given data:
m_solution = mass of water + mass of KBr = 125 g water + 10.5 g KBr = 135.5 g
C_solution = 4.18 J/Cg (given)
ΔT = final temperature - initial temperature = 21.1°C - 24.2°C = -3.1°C

Now, we can calculate q:

q = 135.5 g * 4.18 J/Cg * (-3.1°C)
q = -1727.337 J

Since the dissolving of KBr in water is an exothermic process (heat is released), the value of q will be negative.

To calculate the enthalpy change for dissolving the salt in J/g:

ΔH = q / m_solution
ΔH = -1727.337 J / 135.5 g
ΔH ≈ -12.74 J/g

To convert the enthalpy change to kJ/mol, we need to know the molar mass of KBr. The molar mass of potassium bromide (KBr) is approximately 119 g/mol.

To calculate the enthalpy change for dissolving the salt in kJ/mol:

ΔH = ΔH (J/g) * Molar mass of KBr / 1000
ΔH = -12.74 J/g * 119 g/mol / 1000
ΔH ≈ -1.517 kJ/mol

Therefore, the enthalpy change for dissolving KBr in water is approximately -12.74 J/g and -1.517 kJ/mol.

q = heat released by the water = mass H2O x specific heat H2O x (Tfinal-Tinitial).

Then q/gram KBr = q/10.5 = dH J/g.
dH J/mol = J/gram KBr x (molar mass KBr/1 mol KBr)
Then J/mol x (1 kJ/1000 J) = kJ/mol

Don't forget that this is heat released by the H2O (it gets colder) so the KBr is absorbing q.

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