No. The speed of light does change. You mean speed of light in a vacuum. Measuring that value is elementary, in a 2-way measurement, as you show.
Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.
Explanation : As a single egg cell has an approximately diameter of 100 μm.
We can use this formula to calculate area of the cell membrane;
A = π ;
We can take π as 3.14 and we get;
A = 3.14 X
Soving we get;
A = 7850 μ
Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μ but;
here 1 μ = 7850 μ dye molecules.
Therefore, 10 fluorescent molecules will need;
7850 X 10 = 78500 molecules of dye.
Therefore, the answer is 78500 molecules of dye.
You have to figure out a way to write the two unknown abundances in terms of one variable.
The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).
203(X) + 205(1-X) = 204.4
Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.
After that the higher percentage would be the most abundant.
203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6
x = 0.3
1-x = 0.7
Then the TI-205 would have the highest percentage and would be the most abundant.
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.