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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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B. Direction: In each of the following situations, identify the method of heat transfer that takes place
Ainat [17]

Answer:

1) Conduction

2)Covection

3)Radiation

4)Convection (Land breeze one of the application of convection of heat)

5) Convection

6)Radiation

7) Radiation

8)Conduction

9) Conduction

10) Radiation

Hope it helps

4 0
3 years ago
How would the moon appear from Earth if the moon did not rotate?
andrey2020 [161]
It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon
6 0
3 years ago
Read 2 more answers
A protein molecule in an electrophoresis gel has a negative charge.The exact charge depends on the pH of the solution, but 30 ex
Ad libitum [116K]

Answer:

The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

Explanation:

Given that,

Electric field = 1500 N/C

Charge = 30 e

We need to calculate the magnitude of the electric force on a protein with this charge

Using formula of electrostatic force

F=Eq

Where, F = force

E = electric field

q = charge

Put the value into the formula

F=1500\times30\times1.6\times10^{-19}

F=7.2\times10^{-15}\ N

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6 0
3 years ago
What is the length of 0.005 mm in kilometers?
ZanzabumX [31]
The answer would be 5*10^9
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If we divide \frac{5*10^9}{1*10^6}, it gives us 0.005!

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4 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
3 years ago
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