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Gala2k [10]
3 years ago
7

Are the noble gases considered inert or reactive? Why?

Chemistry
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

Noble gases are considered as Inert because they are unreactive.

Explanation:

Seven noble elments are known to us , they are: helium, neon, argon, krypton, xenon, radon, and oganesson. Noble gases have 8 electrons in their outermost shell and hence their valence shell is complete, they donot need further electrons to complete their octet so they are unreactive. Also they posses following properties.

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Give the clarification of heat and temperature on the basis of molecular motion?
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Explanation:

Heat is a form of thermal energy.

Heat is the sum of all the energy of the molecular motion in an object.

Temperature measures the average heat possessed by each molecule in a given substance.

 Molecules at a higher temperature possess more kinetic energy and they will move faster. This kinetic energy form is the heat variant of thermal energy.

Temperature is the measure of this heat energy of molecules.

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Why is ice in a glacer considered to be a mineral but water a glacier is not
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If you have a book, read it!! I promise you, it tells you this answer!
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Why is making a peanut pretzel and cereal mixture
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How many moles of nitrogen, N, are in 61.0 g of nitrous oxide, N2O?
Oksana_A [137]

Answer:

2.77 mol N

Explanation:

M(N2O) = 2*14 + 16 = 44 g/mol

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                    1 mol         2 mol

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4 0
3 years ago
What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when v
ad-work [718]

Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is \lambda.

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     Optical path length through benzene (\lambda) = 2 \times n \times d

Hence, substituting the given values into the above formula as follows.

    Optical path length through benzene = 2 \times n \times d

                   d = \frac{\text{Optical path length through benzene}}{2 \times n}

                       = \frac{\lambda}{2 \times n}  

                       = \frac{615 \times 10^{-9}}{2 \times 1.501}   (as 1 nm = 10^{-9}m

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Thus, we can conclude that minimum thickness of benzene is 204.9 m.

4 0
3 years ago
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