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Luba_88 [7]
3 years ago
10

How many electrons are depicted in the electron dot diagram of an electrically neutral nitrogen atom? A. two B. six C. eight D.

five
Physics
2 answers:
ella [17]3 years ago
8 0
The amount of electron that are depicted in the electron dot diagram of an electrically neutral nitrogen atom is : D. Five
In ELectron dot diagram, the one that being showed is only the outer energy level

Hope this helps
goldenfox [79]3 years ago
4 0

Answer is D - five.


<em>Explanation;</em>


- Electron dot diagrams show the valence electrons around the element by using dots.


- Valence electrons are the electrons which are in outermost shell of the atom.


-The atomic number of the N atom is 7.

      Atomic number = number of protons = 7

  If the atom is neutral,

      number of protons = number of electrons.


  Hence, N atom has 7 electrons.


- The electron configuration is 1s² 2s² 2p³.


Hence, N atom has 2 + 3 = 5 valence electrons. So, five electrons are represented in electron dot diagram of N.

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In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir
Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be S_{1}

and the distance covered by Sir Alfred be S_{2}

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

S = ut + \frac{1}{2}at^{2}

Where S is the distance traveled

u is the initial velocity

a is the acceleration

and t is the time

For Sir George,

S = S_{1}

u = 0 m/s (Since they start from rest)

a =0.21 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{1}  = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1}  = 0.105 t^{2}\\

t^{2} = \frac{S_{1}}{0.105}

Now, for Sir Alfred

S = S_{2}

u = 0 m/s (Since they start from rest)

a =0.26 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{2}  = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2}  = 0.13 t^{2}\\

t^{2} = \frac{S_{2}}{0.13}

Since, they traveled for the same time, t just before collision, we can write

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

S_{1} + S_{2} = 86 m

Then, S_{2} = 86 - S_{1}

Then,

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13} becomes

\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}

0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}

S_{1} = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

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