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Finger [1]
2 years ago
5

If a concentration of 10 fluorescent molecules per μm2of cell membrane is needed to visualize a cell under the microscope, how m

any molecules of dye would be needed to visualize a single egg? Enter your answer in scientific notation rounded to two significant figures.
Chemistry
1 answer:
lara [203]2 years ago
6 0

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 (100)^{2} / 4;  

We can take π as 3.14 and we get;

A = 3.14 X (100)^{2} / 4  

Soving we get;

A =  7850 μm^{2}  

Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μm^{2}  but;

here 1 μm^{2} = 7850 μm^{2} dye molecules.

Therefore, 10 fluorescent molecules will need;  

7850 X 10 = 78500 molecules of dye.

Therefore, the answer is 78500 molecules of dye.

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2 years ago
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Explain and illustrate the notation for distinguishing between the different p orbitals in a sublevel.
masha68 [24]

Here, the three different notation of the p-orbital in different sub-level have to generate

The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number m_{l} depends upon the value of l, which are -l to +l.

Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as p_{x}, p_{y} and p_{z} in space.

The three p-orbital can be distinguish by the quantum numbers as-

For 2p orbitals (principal quantum number is 2)

1) n = 2, l = 1, m = -1

2) n = 2, l = 1, m = 0

3) n = 2, l = 1, m = +1

Thus the notation of different p-orbitals in the sub level are determined.  

6 0
1 year ago
What type of orbitals overlap to provide stability to the tert-butyl carbocation by hyperconjugation
larisa86 [58]

Explanation:

The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.

Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.

7 0
2 years ago
A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

        \frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}

           V₂ = 124.91mL

6 0
2 years ago
A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the all
blondinia [14]

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

7 0
2 years ago
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