The amount of 2.00 mol of a solid ionic compound was dissolved in 225 mL of water in an insulated container. The initial temp of water was 20 deg. C and the final temp. was 47.8 deg C. What is the delta heat Dissolution? Assume the heat absorbed by the container is negligible.
d=m/v
v=225mL=225cm³
d(water)=1g/cm³
m=225g
Q=mcΔT
Q=225 × 4200×(47.8 –20)
Q=225 × 4.2 × 27.8
Q=–26271j/mol
Q=–26.271kj/mol
2mole ==> –26.271kj/mol
1mole==> xkj/mol
x=–13.1355kj/mol
heat = q = mass water x specific heat water x (Tfinal-Tinitial).
That is the q/2 mol.
This was a multiple choice question, with possible answers of -13.1 KJ/mol; -6.26 KJ/mol; -27.8 KJ/mol; or -26.2 KJ/mol.
I was confused as to how to set equation but now I'm confused because if I'm doing this right-none of these answers work.
Thanks!
For anyone in the future, this is because you have q=mcat is solving for Joules, so convert that to kJ and divide by 2 to get kJ/mol.
Well, well, well, it seems we have a chemistry question here. Let me summon my clown chemistry skills to answer this one for you.
To calculate the delta heat of dissolution, we can use the equation:
q = m × C × ΔT,
where q is the heat absorbed or released, m is the mass of the compound in grams, C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature in degrees Celsius.
But wait a minute, we don't have the mass of the compound just yet. To find that, we need to use the molar mass of the compound and the number of moles given. So, let me do some quick calculations behind my clownish mask.
The molar mass of the compound is not given, so I'll have to use my magical imagination here. Let's pretend the molar mass is 100 g/mol for simplicity.
Now, 2.00 mol of the compound is dissolved. So, the mass of the compound would be:
mass = moles × molar mass
mass = 2.00 mol × 100 g/mol
mass = 200 g.
Now we have the mass of the compound, 200 grams. Let's plug it into our equation. Are you ready for the grand reveal?
q = 200 g × 4.18 J/g°C × (47.8°C - 20°C).
Calculating this, we get:
q ≈ 199,424 J (or about 199.4 kJ).
So, the delta heat of dissolution would be approximately 199.4 kilojoules. Keep in mind, though, that this calculation depends on the molar mass assumption. So, take my answer with a grain of clownish salt, and double-check everything to make sure I haven't pulled any tricks on you.
To calculate the heat of dissolution (delta H_diss) in this scenario, we can use the equation:
delta H_diss = (mass of water) x (specific heat capacity of water) x (change in temperature)
First, we need to calculate the mass of water. We know that the volume of water is 225 mL, but we need the mass in grams. To do this, we can use the density of water, which is approximately 1 g/mL. Therefore:
mass of water = volume of water x density of water
mass of water = 225 mL x 1 g/mL
mass of water = 225 g
Next, we need the specific heat capacity of water, which is approximately 4.18 J/g·°C.
Now we can calculate the change in temperature:
change in temperature = final temperature - initial temperature
change in temperature = 47.8°C - 20.0°C
change in temperature = 27.8°C
Finally, we can substitute these values into the equation for delta H_diss:
delta H_diss = (mass of water) x (specific heat capacity of water) x (change in temperature)
delta H_diss = 225 g x 4.18 J/g·°C x 27.8°C
By multiplying these values together, you can find the delta H_diss.