A- Identify the mixture:
The mixture of powdered charcoal and powdered sugar is considered as a homogeneous mixture. This means that you cannot identify the components with naked eye as they are uniformly distributed in the mixture.
B- Separate components:
You ca separate the charcoal powder from the sugar powder using the following steps:
1- add water. Sugar will dissolve in water while charcoal won't.
2- filter the solution where the powdered charcoal will remain on the filter paper and the solution of powder will pass through.
3- boil the sugar solution (above 100 degrees celcius). The water will evaporate and the sugar will precipitate.
Climate is considered an abiotic limiting factor because it is non living . hope this helped :))
From the equation q=mCΔT, set the q of copper = to q of water,
So --- mCΔT(copper)=mCΔT(water).
mass (Cu - copper) = 38g
mass (H2O - water) = 15g
C (H2O) = 4.184 J/g*C
ΔΤ (H2O) = 33-22 = 11*C
ΔΤ (Cu) = 33-80 = -47*C (the final temp is the same for both materials - thermal equilibrium)
C (Cu) = ?
So --- 38(-47)C[Cu]=15(4.184)(11)
--- C[Cu]=690.36/(-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)
Answer: I would prepare a standard solution
Explanation: A standard solution is a solution of known concentration (molarity). This is gotten from the stock using dilution principle. (C1V1=C2V2) I would prepare a 0.1M solution of the beverage into a standard 500ml flask.
