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padilas [110]
3 years ago
14

_ method is used to remove soluble impurities. ​

Physics
2 answers:
Novosadov [1.4K]3 years ago
7 0

Answer:

To remove soluble impurities, first, by doing solubility tests, a suitable solvent is chosen (high solubility in hot solvent, low solubility in cold solvent). The soluble impurities are then removed as follows: the desired compound along with the soluble impurities are dissolved in a MINIMUM of near-BOILING solvent.

anygoal [31]3 years ago
5 0

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Methods of Separation.

Evaporation and Decantation are the process which can be easily used to remove the Soluble Impurities.

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Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe
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Answer:

Obviously Lengthen...   T = 2\pi \sqrt{L/g}   or   g = 4\pi ^{2} L/g

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

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3 years ago
Explain why greenhouse gases in the atmosphere cause an increase in temperature?
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3 years ago
Where do you see triangulation used on this structure? Explain how triangulation​
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3 0
3 years ago
20. Consider a model steel bridge that is 1/100 the exact scale of the real bridge that is to be built. a. If the model bridge w
Veseljchak [2.6K]
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.

Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.

Because the model bridge weighs 50 N, the real bridge weighs
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6 0
4 years ago
You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength
Ray Of Light [21]

Answer:

wavelength \lambda = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = \frac{D\lambda}{d}    ....................1

\lambda is Wavelength of light and  D is Distance between screen and slit and d is slit width

so put here value and we get

\lambda = \frac{2.96*0.325*10^{-6}}{2.20}    

\lambda = 437.27 × 10^{-9} m

wavelength \lambda = 437.27 nm

4 0
3 years ago
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