The heat of reaction for burning 1 mole of a certain compound X is known to be -477.7 kJ. The calorimeter constant of the bomb being used is 2500 j/degree celsius and the initial temperature of the water is 23.2 degrees celsius.

a) if 22.54 g of compound X (MM=46.0) is burned in the bomb calorimeter containing 2000. ml of water (S.H.=4.184 j/g(degrees celsius), what will be the final temperature?

b) How much water can be warmed from 23.2 C to 56.5 C when 52.3 g of the compound is burned in the bomb?

22.54/46.0 = moles cmpd X.

heat generated by 22.54 moles = 477.7 kJ/mole x (22.54/46.0) = ?? kJ.
Then heat generated (change to Joules) = mass water x specific heat water x (Tfinal-Tinitial) + 2500 J/C*(Tfinal-Tinitial)
Solve for Tfinal.

The second problem is the same concept.

i still keep messing up, can someone give me a specific answer i can use as a base?

I estimated the answer to the first problem as about 45 C or so.

To find the final temperature in both cases, we can use the principle of conservation of energy. The heat released by the burning compound X will be equal to the heat absorbed by the water and the bomb calorimeter.

a) Before calculating the final temperature, we need to convert the mass of compound X to moles. We can use the molar mass (MM) of compound X for this conversion.

Given:
Mass of compound X (m) = 22.54 g
Molar mass of compound X (MM) = 46.0 g/mol

To find the number of moles (n) of compound X:
n = m / MM
n = 22.54 g / 46.0 g/mol ≈ 0.489 moles

Next, we can calculate the heat released by the burning of compound X:
ΔH = -477.7 kJ (given)

Now, let's convert the calorimeter constant from J/°C to kJ/°C:
Calorimeter constant (C) = 2500 J/°C = 2.5 kJ/°C

The heat absorbed by the water and the bomb calorimeter will be:
Q = n * ΔH + C * ΔT

Substituting the given values:
Q = 0.489 moles * -477.7 kJ/mol + 2.5 kJ/°C * ΔT

Rearranging the equation to solve for ΔT:
ΔT = (Q - n * ΔH) / C

Substituting the values:
ΔT = (0.489 * -477.7) / 2.5

Finally, we can calculate the final temperature (T_final) by adding the calculated ΔT to the initial temperature:
T_final = 23.2°C + ΔT

b) In this case, we'll follow a similar approach to calculate the final temperature. However, we'll need to find the moles of compound X and adjust the heat equation accordingly.

Given:
Mass of compound X (m) = 52.3 g

To find the number of moles (n) of compound X:
n = m / MM
n = 52.3 g / 46.0 g/mol ≈ 1.137 moles

The heat absorbed by the water and the bomb calorimeter will now be:
Q = n * ΔH + C * ΔT

Since the final temperature is not given, we can calculate the change in temperature (ΔT) first using the equation:
ΔT = (Q - n * ΔH) / C

Substituting the given values:
ΔT = (1.137 * -477.7) / 2.5

Finally, the amount of water that can be warmed from 23.2°C to 56.5°C will be determined by its heat capacity.

Given:
Specific heat of water (S.H.) = 4.184 J/g°C
Initial temperature (T_initial) = 23.2°C
Final temperature (T_final) = 56.5°C

We can use the formula:
Q = m * S.H. * ΔT

Rearranging the equation to solve for mass (m):
m = Q / (S.H. * ΔT)

Substituting the values:
m = Q / (4.184 J/g°C * (T_final - T_initial))

To convert the final mass to milliliters, we'll need to know the density of water:
Density of water = 1 g/mL

m (in grams) = calculated mass in grams
Volume (in mL) = calculated mass in grams

Therefore, the amount of water that can be warmed from 23.2°C to 56.5°C is equal to the calculated mass in grams, which is also equal to the volume in milliliters.