Answer:
0.67mol/Kg
Explanation:
The following were obtained from the question:
Mole of solute = 0.50mol
Mass of solvent = 750g = 750/1000 = 0.75Kg
Molality =?
Molality = mole of solute /mass of solvent
Molality = 0.5/0.75
Molality = 0.67mol/Kg
This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:
<h3>Heating curves:</h3>
In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.
Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:
Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:
![Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\ \\ Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\ \\ Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\ \\ Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\ \\ Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ](https://tex.z-dn.net/?f=Q_1%3D0.0217mol%2A111.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28-114.1%5C%C2%B0C%29-%28-200%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.208kJ%5C%5C%0A%5C%5C%0AQ_2%3D0.0217mol%2A4.9%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.106kJ%5C%5C%0A%5C%5C%0AQ_3%3D0.0217mol%2A112.4%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%2878.4%5C%C2%B0C%29-%28-114.1%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.470kJ%5C%5C%0A%5C%5C%0AQ_4%3D0.0217mol%2A38.6%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.838kJ%5C%5C%0A%5C%5C%0AQ_5%3D0.0217mol%2A87.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28150%5C%C2%B0C%29-%2878.4%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.136kJ)
Finally, we add them up to get the result:
Learn more about heating curves: brainly.com/question/10481356
The equilibrium constant is 1.3 considering the reaction as written in the question.
<h3>Equilibrium in chemical reactions</h3>
In a chemical reaction, the equilibrium constant is calculated based on the equilibrium concentration of each specie. The equation of this reaction is;
A (g) + 2B (g) ⇌ 3C (g).
The initial concentration of each specie is;
- A - 9.22 M
- B - 10.11 M
- C - 27.83 M
The equilibrium concentration of B is 18.32 M
We now have to set up the ICE table as follows;
A (g) + 2B (g) ⇌ 3C (g)
I 9.22 10.11 27.83
C -x -x +x
E 9.22 - x 10.11 - x 27.83 + x
The equilibrium concentration of B is 18.32 M hence;
10.11 - x = 18.32
x = 10.11 - 18.32 = -8.21
Hence;
Equilibrium concentration of A = 9.22 - (-8.21) = 17.43
Equilibrium concentration of C = 27.83 + (-8.21) = 19.62
Equilibrium constant K = [19.62]^3/[17.43] [18.32]^2
K = 1.3
Learn more about equilibrium constant: brainly.com/question/17960050
The temperature that water boils on the fahrenheit scale is c. 212°f.
<h3>What is Temperature conversation?</h3>
Temperature conversation is a way of converting a temperature from a unit to another using conversation scale.
Therefore, on fahrenheit scale water boils at 212°f which is equivalent to 100°C on the Celcius scale.
Learn more about Temperature conversation at;brainly.com/question/23419049
#SPJ4
Answer:
1. H=2, O=1
2. H=4, O=2
3. C=1, O=1
4.
a. reactants C=1, O=2
b. products C=1, O=2
5.
a. reactants H=4, O=2
b. products H=4, O=2
6
a. reactants C=1, O=1, H=4
b. products C=1, O=1, H=4
