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liberstina [14]
3 years ago
13

Lithium and fluorine undergo ionic bonding. Using the noble gas electron configurations for each (below), please explain the pro

cess of bonding step by step, using proper grammar and mechanics. Noble Gas Electron Configurations:________ Lithium:_______ [He] 2s1 Fluorine:______ [He] 2s22p5
Chemistry
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

Lithium loses one electron to fluorine and forms ionic bond, having formula LiF.

Explanation:

Lithium is the element of the group 1 and period 2 which means that the valence electronic configuration is [He]2s^1.

Fluorine is the element of the group 17 and period 2 which means that the valence electronic configuration is [He]2s^22p^5.

<u>Thus, lithium loses 1 electron and become positively charged. Fluorine on the other hand accepts this electron and become negatively charged.</u> This is done in order that the octet of the atoms are complete.  <u>These both ions then form ionic bond as their will be electrostatic interaction between the two oppositely charged ions.</u>

Thus, the formula of calcium chloride is LiF.

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IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at 75^{o}F and 14.6 psia.

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(a)  Formula to calculate hydraulic radius (r_{H}) is as follows.

              r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}

                          = \frac{2 \times 1}{2(1) + 2(2)}

                          = \frac{1}{3} ft

Formula for equivalent diameter is as follows.

                     D_{eq} = 4 \times r_{H}

                                    = 4 \times \frac{1}{3} ft  

                                    = \frac{4}{3} ft

(b)    Formula for velocity floe is as follows.

                         Q = VA

                     V = \frac{Q}{A}

                        = \frac{48000}{2 \times 1} ft/min

                        = 24000 ft/min

(c)   Formula to calculate Reynold's number is as follows.

         R_{e} = \frac{D \times V \times \rho}{\mu}

                   = \frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}  (as \rho = 0.0744 lb/ft^{3} and \mu = 0.0443 lb/ft. hr)

                   = 53742.66 hr/min  

As 1 hr = 10 min. So, 53742.66 hr/min \times \frac{60 min}{1 hr}

                            = 3224559.6

(d)   Formula to calculate pressure drop (\Delta P) is as follows.

              \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

Putting the given values into the above formula as follows.

               \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

                      = \frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}

                      = 6.238 lb/ft^{2}

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<u>Correct Question Format:</u>

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