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.
Answer: B. triH sol Mgl2= -triHlat+ triHhydr Mg^2+ 2triHhydr^l-
Explanation:
Just did it and it was right
Answer:
C
Explanation:
75 mile shallow flat area just off coastlines
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
