Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
Answer:
The differemt isotopes that differ in atomic mass
Explanation:
Always use this method !!! always
I believe it's answer #3. Logically, at least.
You can test #1 through trial and error.
You can experiment #2 also through trial and error.
You cannot test #3 through trial and error, because that would be catastrophic.
You can test #4 through a survey and individual study and data collection.
