Answer:
Is this an actual question that you had for school?
Explanation:
Q unknown = -Q H2O
Q H2O= (100g x 4.18 J/g C x (25C - 20C)
Q H2O = 2090J
-2090J = (50g x c x (25C-90C)
-2090J = -3250c
.64 = c
I am a high school student and have done this problem a few months ago for my test, hope this helps
One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.
Precipitation titration is a titration in which a reaction occurs from the analyte and titrant to form an insoluble precipitate.
With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.
The precipitation titrimetry methods with the use of argentometry includes
• Mohr’s Method
• Fajan’s Method
• Volhard’s Method
Difficulties encountered in precipitation titration includes
- Getting the exact end point is hard.
- it is a very slow titration method.
- it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.
See more on Precipitation: brainly.com/question/20628792
From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent.
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K
C2H6 is a chemical compound
