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Alex_Xolod [135]
3 years ago
5

How can parts of a solution be separated by chromatography?

Physics
2 answers:
zysi [14]3 years ago
3 0

The parts of a solution be separated by chromatography as "the solution is absorbed onto a paper".

Answer: Option D

<u>Explanation:</u>

A solution is a homogeneous mix of two substances or more. When chemists have to evaluate what components in a solution or other mixture are available, they often use a method called chromatography.

One easy way to determine how chromatography extracts the components of a solution is to look at what happens when a scrap of paper that is written on it becomes wet. Paper chromatography is an experimental tool used to distinguish compounds or colored chemicals.

lutik1710 [3]3 years ago
3 0

Answer:

D

Explanation:

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Answer:

W_2=-12J

Explanation:

The work of force 2 will be given by the vectorial equation W_2=F_2.d. We know the value of F_1 and have information about its movement, which relates to the net force F=F_1+F_2.

About this movement we can obtain the acceleration using the equation v_f^2=v_i^2+2ad. Since it departs from rest we have a=\frac{v_f^2}{2d}.

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1

And we had the work done by force 2 as:

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(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J

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<em>W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}</em>

<em>Which leads us to the previous equation straightforwardly.</em>

6 0
3 years ago
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how
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Answer:

W =  343.2 J

Explanation:

Given that,

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A very low value of the equilibrium constant for a reaction can indicate that
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That reactants are favored.
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The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is suf
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Answer:

(a) I_{1}=3.2*10^{-3}W/m^{2}

(b) \beta =95dB

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2}  }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2}  }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2}  }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta  =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2}  }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB

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Explanation:

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