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3 years ago

6

What is the method use to separate sand and water

1 answer:

Slav-nsk [51]3 years ago

7 0

Sand and water can be separated by any of the following methods:

1. Sedimentation and decantation: This method involves the mixture being kept undisturbed for some time. After some time, sand being heavier and insoluble in water, settles down at the bottom of container. Now, water is poured into another container to separate it from sand.

2. Filtration: This method involves the mixture being passed through a filter paper (a filter with very fine pores). Sand particles being larger in size are retained by the filter paper and get separated from water.

I hope this helps! :D

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In a first–order reaction (A → B and rate = k [A]), tripling the concentration of A will have what effect on the reaction rate?

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3 years ago

A balloon has 18.9 moles of hydrogen gas at a volume of 428 L. What would the volume of the gas be if the balloon had 14.0 moles

den301095 [7]

Answer:

V₂ = 317 L

Explanation:

Given data:

Initial number of moles of hydrogen = 18.9 mol

Initial volume of gas = 428 L

Final volume = ?

Final number of moles = 14.0 mol

Solution:

According to the Avogadro law,

Number of moles of gas is directly proportional to the volume of gas at constant temperature and pressure.

Mathematical expression:

V₁/n₁ = V₂/n₂

V₁ = Initial Volume of balloon

n₁ = initial number of moles

V₂ = Final volume of balloon

n₂ = Final number of moles

Now we will put the values.

428 L / 18.9 mol = V₂/ 14 mol

V₂ =428 L × 14 mol / 18.9 mol

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4 years ago

How many feet will you travel in 1.5 hours if you drive at a speed of 156 mph

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3 years ago

What is the wavelength in picometers for an electron with 360 eV of kinetic energy?

Rom4ik [11]

Answer:

Wavelength = 64.635 pm

Explanation:

The expression for the deBroglie wavelength and kinetic energy is:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

Where,

$\lambda$ is the deBroglie wavelength

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

K.E. is the kinetic energy of the electron.

Given, K.E. = 360 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, K.E. = $360\times 1.6022\times 10^{-19}\ J=5.76792\times 10^{-17}\ J$

Applying in the equation as:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

$\lambda=\frac{6.626\times 10^{-34}}{\sqrt {{2\times 9.11\times 10^{-31}\times 5.76792\times 10^{-17}}}}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{\sqrt{10^{-48}\times \:105.0915024}}\ m$

$\lambda=6.4635\times 10^{-11}\ m$

Also, 1 m = 10¹² pm

<u>So, Wavelength = 64.635 pm</u>

4 0

4 years ago