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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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With a 30 mph head wind it takes the plane 18.52 hours to fly 5000 miles. ANSWER 2: With a 30 mph tail wind it takes the plane 15.15 hours to fly 5000 miles.
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3 years ago
Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average r
Dvinal [7]

Answer:

The lifetime of the particle is  \Delta  t  =  2.6*10^{-25} \ s

Explanation:

From the question we are told that

    The average rest energy is E =  91.19 \ GeV =  91.19GeV  *    \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J

    The intrinsic width is  \Delta E  =2.5eV  = 2.5GeV   *  \frac{1.60 *10^{-10}J  }{1GeV}  =  4*10^{-10} J

The lifetime is mathematically represented as

     \Delta  t  =  \frac{h}{\Delta E}

Where h is the Planck's constant with a value of  1.055*10^{-34} \ J\cdot s  

substituting values

    \Delta  t  =  \frac{1.055*10^{-34}}{4 *10^{-10}}

     \Delta  t  =  2.6*10^{-25} \ s

6 0
3 years ago
This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon
adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

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Hope this helped!

8 0
3 years ago
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.
Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}

a = 30 units    and  b = 70 units

= \sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

                       

4 0
3 years ago
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