20.0 g of ammonium nitrate (NH4NO3) is dissolved in 125 g of water in a coffee-cup calorimeter, the temperature falls from 296.5 K to 286.4 K. Find the value of q for the calorimeter.

Well, it seems like you've got quite the chemical conundrum there! Let's see if we can unravel it.

To find the value of q for the calorimeter, we can use the equation:

q = m × c × ΔT

Where:
q is the heat transfer
m is the mass
c is the specific heat capacity
ΔT is the change in temperature

Now, let's break it down step by step.

First, we need to find the mass of the solution. The total mass will be the sum of the mass of the ammonium nitrate (20.0 g) and the mass of water (125 g). So, the mass (m) would be 20.0 g + 125 g = 145.0 g.

Next, we need to find the specific heat capacity (c) of the solution. Since the solution is primarily water, we can assume that the specific heat capacity of the solution is approximately the same as that of water (4.18 J/g·K).

Finally, we need to find the change in temperature (ΔT). The initial temperature (Ti) is 296.5 K and the final temperature (Tf) is 286.4 K. So, the change in temperature (ΔT) would be 286.4 K - 296.5 K = -10.1 K.

Now, let's plug in the values into the equation and calculate the value of q:

q = m × c × ΔT
= 145.0 g × 4.18 J/g·K × (-10.1 K)
= -6090.95 J

So, the value of q for the calorimeter is approximately -6090.95 J.

But hey, cheer up! At least we've given your brain a little workout with this math problem. Remember, laughter is the best calorie-burner!

To find the value of q for the calorimeter, we can use the equation:

q = m * C * ΔT

Where:
q = heat absorbed or released by the system
m = mass of the solution (125 g of water)
C = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)

Given:
Mass of solution (water) = 125 g
Specific heat capacity of water (C) = 4.18 J/g°C
Initial temperature (Ti) = 296.5 K
Final temperature (Tf) = 286.4 K

First, let's calculate the change in temperature (ΔT):

ΔT = Tf - Ti
ΔT = 286.4 K - 296.5 K
ΔT = -10.1 K

Now we can calculate the value of q:

q = m * C * ΔT
q = 125 g * 4.18 J/g°C * (-10.1 K)
q = -5272.5 J

The value of q for the calorimeter is approximately -5272.5 Joules. Since the temperature falls, the negative sign indicates that heat is released by the system, and the calorimeter absorbs it.

To find the value of q for the calorimeter, we can use the equation:

q = m * c * ΔT

where:
q is the heat transfer (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·K),
ΔT is the change in temperature (in Kelvin).

First, we need to calculate the heat transfer for the ammonium nitrate (NH4NO3).

Given:
Mass of NH4NO3 (m) = 20.0 g
Change in temperature (ΔT) = 296.5 K - 286.4 K = 10.1 K

Next, we need to find the specific heat capacity of NH4NO3. However, the specific heat capacity of ammonium nitrate is not provided. Therefore, we need to look up or calculate its value.

The specific heat capacity of ammonium nitrate can be found in reference books or online sources. For reference, the specific heat capacity of several common substances is:

- Water (c) = 4.18 J/g·K
- Ammonium nitrate (NH4NO3) is approximately 2.18 J/g·K

With the specific heat capacity value, we can now calculate the heat transfer for the ammonium nitrate:

q_nh4no3 = m * c_nh4no3 * ΔT
= 20.0 g * 2.18 J/g·K * 10.1 K
= 441.16 J

Now, let's calculate the heat transfer for the water present in the calorimeter:

Given:
Mass of water (m) = 125 g
Change in temperature (ΔT) = 296.5 K - 286.4 K = 10.1 K
Specific heat capacity of water (c) = 4.18 J/g·K

q_water = m * c * ΔT
= 125 g * 4.18 J/g·K * 10.1 K
= 5279.25 J

To find the total heat transfer in the calorimeter, we sum up the heat transfers for NH4NO3 and water:

q_total = q_nh4no3 + q_water
= 441.16 J + 5279.25 J
= 5720.41 J

Therefore, the value of q for the calorimeter is approximately 5720.41 J.

When NH4NO3 is dissolved in water, it dissociates into its ions, followed by a temperature drop:

NH4NO3 -> NH4+ + NO3-
(though it's not really that relevant though ^^;)

Anyway, the heat gained or lost can be calculated using the formula
Q = mc(T2 - T1)
where
m = mass (grams)
c = specific heat capacity (J/g-K)
T = temperature (K)

Here, we'll use an assumption that the specific heat of the solution is equal to that of water, which is c = 4.184 J/g-K.
The total mass of solution is 20 + 125 = 145 grams.
Substituting,
Q = (145)*(4.184)*(286.4 - 296.5)
Q = -6127.47 J
The Q negative because heat is released in the dissociation.

Check the significant figures.
Hope this helps :3