The lower the percent of energy required for a machine to perform work, the more energy efficient it is. Please select the best
2 answers:
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Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) =
where; solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6= ÷
4.6 =
4.6 × =
4.6 B = 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B =
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.
Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
<em></em>
Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere =
==> V = = 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>