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1 year ago

13

Now explain your diagnosis. Start your argument by writing something like this: "My group believes that Elisa has/does not have

___. I think that she does/does not have the ___ condition because . . ." (Then, explain how molecules move through the body when someone has the condition you investigated, and compare that to Elisa's test results.) plz help me ;(

1 answer:

likoan [24]1 year ago

4 0

Answer:

Explanation:I would need more info to understand this question but explaining molecules is pretty easy tho

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

lara [203]

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 μ $m^{2}$

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

here 1 μ $m^{2}$ = 7850 μ $m^{2}$ dye molecules.

Therefore, 10 fluorescent molecules will need;

7850 X 10 = 78500 molecules of dye.

Therefore, the answer is 78500 molecules of dye.

6 0

2 years ago

Room temperature is about 20 degrees Celsius. Explain how you could convert this temperature to kelvin. Use evidence from the fi

Taya2010 [7]

Answer:

293.15 K.

Explanation:

It is given that, the room temperature is 20 degrees Celsius.

We need to convert this temperature into kelvin.

The conversion from degrees Celsius to Kelvin is as follows :

$T_k=T_c+273.15$

We have, $T_c=20^{\circ} C$

So,

$T_k=20+273.15\\\\T_k=293.15\ K$

So, the room temperature is 293.15 kelvin.

8 0

2 years ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

2 years ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago

Complete the equation for the dissociation of cdcl2(aq). omit water from the equation because it is understood to be present.

mixas84 [53]

Cadmium chloride is a highly soluble compound. The equation for its dissolution is:

CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.

3 0

2 years ago