Answer:
-1.71 J/K
Explanation:
To solve this problem we use the formula
ΔS = n*ΔH/T
Where n is mol, ΔH is enthalpy and T is temperature.
ΔH and T are already given by the problem, so now we calculate n:
Molar Mass C₂H₅OH = 46 g/mol
2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol
Now we calculate ΔS:
ΔS = 0.0589 mol * −4600 J/mol / 158.7 K
ΔS = -1.71 J/K
Answer: I believe the 1st and 3rd reactions are better obtained through reference sources and the 2nd and 4th are easiest and safest to measure in the laboratory.
Explanation:
I am also working on this Pre-lab right now, and I looked back at the first question to help get my answer. In the first question (a), it is noted that ammonia gas and gaseous hydrochloric acid are both potentially dangerous in gaseous form. I saw that both the 1st and 3rd reactions contained noxious gases (I knew this because there was a (g) in both of these reactions). Using the knowledge from the first question that the noxious gases were potentially dangerous, I assumed that those reactions were the ones that are better obtained through the reference sources. The 2nd and 4th reactions did not contain any noxious gases, so I assumed those ones were easiest and safest to measure in the laboratory. Hope this helps!
Answer:
-0.93 °C
Explanation:
Hello,
The freezing-point depression is given by:
Whereas 
is the freezing temperature of the solution, 
is the freezing temperature of the pure solvent (0 °C since it is water), 
the Van't Hoff factor (1 since the solute is covalent), 
the solvent's freezing point depression point constant (in this case 
) and 
the molality of the glucose.
As long as the unknown is , solving for it:
Best regards.
Well you could travel around the world and discover new kind's of weather! am I right?
