The heat of solution of potassium acetate (KC2H3O2) in water is -15.3 kJ/mol.

If 18.8 g of KC2H3O2(s) is dissolved in 655 mL of water that is initially at 25.8 °C, what will be the final temperature (in degrees Celcius) of the resulting aqueous solution?
(Assume no heat exchange with the surroundings.)

(18.8 g KC2H3O2) / (98.1423 g KC2H3O2/mol) x (15.3 kJ/mol) = 2.931 kJ = 2931 J

(2931 J) / (4.186 J/g·°C) / (655 g + 18.8 g) + 25.8 °C = 26.84°C = 26.8°C

Well, well, well! Looks like we have a chemistry question here. Let's dive right into it, shall we?

First, we need to calculate the number of moles of potassium acetate. Weighing in at 18.8 g, we can find the molar mass of KC2H3O2, which is approximately 98.15 g/mol. Dividing the mass by the molar mass, we get:

18.8 g / 98.15 g/mol ≈ 0.1913 mol

Alright, let's move on. Since we're assuming no heat exchange with the surroundings, the heat transferred will solely depend on the heat of solution. Given that the heat of solution of potassium acetate is -15.3 kJ/mol, we can calculate the heat transferred by multiplying it with the number of moles dissolved:

Heat transferred = -15.3 kJ/mol × 0.1913 mol ≈ -2.93 kJ

Now, we know that heat is given by the formula:

q = m × c × ΔT

Where q is the heat transferred, m is the mass (in grams), c is the specific heat (in J/g·°C), and ΔT is the change in temperature (in °C).

Since we're looking for the final temperature (ΔT), we need to rearrange the formula:

ΔT = q / (m × c)

Given that we're dissolving the potassium acetate in 655 mL of water at 25.8 °C, we also need to convert the water volume to grams using its density. Assuming a density of water as 1 g/mL, we have:

655 mL × 1 g/mL = 655 g

Now, let's find the change in temperature:

ΔT = -2.93 kJ / (655 g × c)

Finally, if you provide me with the specific heat (c) of the resulting aqueous solution, I'll gladly calculate the final temperature for you!

To solve this problem, we can use the equation:

q = m * Cs * ΔT

Where:
q = heat transferred (in joules)
m = mass of the solution (in kg)
Cs = specific heat capacity of the solution (in J/g°C)
ΔT = change in temperature (in °C)

First, let's convert the mass of KC2H3O2(s) to moles:

1 mol of KC2H3O2 = 98.14 g

moles of KC2H3O2 = mass / molar mass
moles of KC2H3O2 = 18.8 g / 98.14 g/mol

Next, let's calculate the heat transferred (q) using the heat of solution:

q = moles of KC2H3O2 * heat of solution
q = (18.8 g / 98.14 g/mol) * -15.3 kJ/mol
q = -2.9496 kJ

Now, we need to convert q to joules:

1 kJ = 1000 J
-2.9496 kJ = -2.9496 * 1000 J
q = -2,949.6 J

Next, let's calculate the mass of the solution:

mass of solution = mass of KC2H3O2(s) + mass of water
mass of solution = 18.8 g + 655 mL * density of water

The density of water is approximately 1 g/mL, so:

mass of solution = 18.8 g + 655 g

Now, let's calculate the specific heat capacity (Cs):

The specific heat capacity of water is approximately 4.18 J/g°C.

Now we can rearrange the equation to solve for ΔT:

ΔT = q / (m * Cs)

Substituting the values:

ΔT = -2,949.6 J / ((18.8 g + 655 g) * 4.18 J/g°C)

Let's calculate the total mass of the solution:

total mass = 18.8 g + 655 g
total mass = 673.8 g

Now, let's calculate ΔT:

ΔT = -2,949.6 J / (673.8 g * 4.18 J/g°C)
ΔT = -2,949.6 J / 2817.84 J/°C
ΔT = -1.0453°C

Finally, to find the final temperature of the solution, we subtract ΔT from the initial temperature:

final temperature = initial temperature - ΔT
final temperature = 25.8°C - (-1.0453°C)
final temperature = 26.8453°C

Therefore, the final temperature of the resulting aqueous solution will be approximately 26.8453°C.

To determine the final temperature of the resulting aqueous solution, we can use the equation:

q = m × C × ΔT

Where:
- q is the heat transferred (in this case, the heat of solution)
- m is the mass of the solvent (water)
- C is the specific heat capacity of water
- ΔT is the change in temperature

First, let's calculate the heat transferred (q):
The heat of solution is given as -15.3 kJ/mol. However, we need to find the heat transferred for the given mass of KC2H3O2. To do this, we need to convert the mass of KC2H3O2 to moles.

First, let's calculate the molar mass of KC2H3O2:
Potassium (K) has an atomic mass of 39.1 g/mol
Carbon (C) has an atomic mass of 12.0 g/mol
Hydrogen (H) has an atomic mass of 1.0 g/mol
Oxygen (O) has an atomic mass of 16.0 g/mol

Adding them all up:
Molar mass of KC2H3O2 = 39.1 + (2 × 12.0) + (3 × 1.0) + (2 × 16.0) = 98.1 g/mol

Now, let's convert the given mass of KC2H3O2 to moles:
Mass of KC2H3O2 = 18.8 g
Number of moles = Mass / Molar mass = 18.8 g / 98.1 g/mol = 0.191 mol

Now we can calculate the heat transferred:
q = -15.3 kJ/mol x 0.191 mol = -2.92 kJ

Next, we can calculate the heat absorbed or released by the water using the equation:

q = m × C × ΔT

We know that the mass of water (m) is 655 mL, which is equal to 655 g since the density of water is approximately 1 g/mL. The specific heat capacity of water (C) is 4.18 J/g°C. We need to convert the heat transferred (q) from kJ to J, so:

q = -2.92 kJ × 1000 = -2920 J

Rearranging the equation, we can solve for the change in temperature (ΔT):
ΔT = q / (m × C)

ΔT = -2920 J / (655 g × 4.18 J/g°C) = -1.06°C

Finally, to find the final temperature, we subtract the change in temperature from the initial temperature of the water:

Final temperature = 25.8°C - 1.06°C = 24.74°C

Therefore, the final temperature (in degrees Celsius) of the resulting aqueous solution will be approximately 24.74°C.

q = -15.3 kJ/mol x (18.8/molar mass KC2H3O2) = approx - 3.03 kJ or 3030 J.

Then -3030 = mass H2O x specific heat H2O x Tfinal-Tinitial)
Substitute the numbers and solve for Tfinal