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shepuryov [24]
3 years ago
15

How many moles are in 5.0 x 10 *8 atoms of carbon

Chemistry
1 answer:
valentina_108 [34]3 years ago
8 0
1 mole of any ... contains 6.02x10^23 atom (Avogadro constant)
so do the revers process of it, what I mean by that is
 6.02xa0^23 atom is present in 1 mole of...
so 5.0*10^8 atom is present in...
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Which of these mixtures will MOST LIKELY result in soil? A) sand, clay, and heat B) sand, clay, and water C) sand, clay, heat, a
Wewaii [24]

Answer:

I think it should be C.

Explanation:

Because for a plant to start photosinthesis it need sun (heat is like the sun),water,and stuff that soil is made of (sand and clay) tell me if this is correct

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5 0
4 years ago
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The first and Third graph
6 0
3 years ago
How many core electrons are in a ground state atom of selenium?
Harman [31]
Electrons in an atom can be classified as core electrons and valence electrons. Valence electrons are those electrons which are present in valence shell and participates in bond formation. While, Core electrons are all remaining electrons which are not present in valence shell, hence not take part in bonding.

Atomic number of Selenium (Se) is 34 hence it has 34 electrons with following electronic configuration;

                          1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴

From electronic configuration it is found that the valence shell is 4, and the number of electrons present in valence shell are 6. So, 

             Core Electrons  =  Total Electrons - Valence Electrons

             Core Electrons  =  34 - 6

             Core Electrons  =  28

Result:
          There are 28 core electrons in Selenium.
3 0
3 years ago
An evil villain wants to broadcast a message to everyone on Earth. They are using satellites for the communication process. For
stira [4]
Uhhhhhhh maybe the answer is 3
7 0
2 years ago
34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i
nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

          p(r) = Ar(r^{2}e^{\frac{-r}{3a_{o}}}

                = Ar^{3}e^{\frac{-r}{3a_{o}}}

     \frac{dp(r)}{dr} = 0

            Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}} + Ae^{\frac{-r}{3a_{o}}}(3r^{2}) = 0

                     \frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}

                      = 3Ar^{2}e^{\frac{-r}{3a_{o}}}

                       r = 9a_{o}

Thus, we can conclude that most likely radius at which the electron would be found is 9a_{o}.

4 0
3 years ago
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