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Gekata [30.6K]
3 years ago
12

Fill in the blank with the appropriate numbers for both electrons and bonds(considering that single bonds are counted as one, do

uble bonds as two, and triple bonds as three).12345678
1. Fluorine has __________ valence electrons and makes __________ bond(s) in compounds.2 Oxygen has __________ valence electrons and makes __________ bond (s) in compounds.3. Nitrogen has __________ valence electrons and makes __________ bond(s) in compounds.4. Carbon has __________ valence electrons and makes __________ bond(s) in compounds.
Physics
1 answer:
tekilochka [14]3 years ago
3 0

Answer:

1.Fluorine is having 7 number of electrons and 1 makes bond.

     Electronic configuration  -  1S² 2S² 2P⁵  

    1 electron need to get in stable states 2P⁶.

2.Oxygen is having 6 balance electron and 2 makes bonds.

    Electronic configuration  -  1S² 2S² 2P⁴  

   2 electron need to get in stable states 2P⁶.

3.Nitrogen is having 5 balance electron and 3 makes bonds.

    Electronic configuration  -  1S² 2S² 2P³

    3 electron need to get in stable states 2P⁶.

4. Carbon having 4 balance electron and 4  makes bonds.

    Electronic configuration  -  1S² 2S² 2P²

     4 electron need to get in stable states 2P⁶.

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5 0
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Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
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According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



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