Answer:
Try this link https://chem.libretexts.org/Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_(Exercises)
Answer:
%KCl = 7.05%
%Water = 92.95%
Explanation:
Step 1: Given data
- Mass of KCl (solute): 36 g
- Mass of water (solvent): 475 g
Step 2: Calculate the mass of the solution
The mass of the solution is equal to the sum of the masses of the solute and the solvent.
m = 36 g + 475 g = 511 g
Step 3: Calculate the mass percentage of the solution
We will use the following expression.
%Component = mComponent/mSolution × 100%
%KCl = 36 g/511 g × 100% = 7.05%
%Water = 475 g/511 g × 100% = 92.95%
Answer:
Crushing a can.
Melting an ice cube.
Boiling water.
Mixing sand and water.
Breaking a glass.
Dissolving sugar and water.
Shredding paper.
Chopping wood.
Answer:
1672 J
Explanation:
We are supposed to find out the energy released by water when cooled from 50° C to 30°C .
We have the relation that ,
H=msΔT
Information is given that s=4.18 J/ g*°C ,
m=20.0 g,
and ΔT is -20° C.
Hence by substituting we get,
H=-20*20*4.18
(minus symbol represents that energy is being released)
1672 J is release when water is cooled from 50° C to 30°C .
Answer:
38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) =
21.6 g Cl in C2F3Cl3
