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Andrei [34K]
2 years ago
8

What best explains whether bromine (Br) or neon (Ne) is more likely to form a covalent bond? On left, a purple circle labeled Br

surrounded by 4 concentric circles. The inner circle has 2 small green spheres. The second circle has 8 small green spheres. The third circle has 18 small green spheres. The fourth circle has 5 small green spheres. On right, a purple circle labeled Ne surrounded by 3 concentric circles. The inner circle has 2 small green spheres. The middle circle has 8 small green spheres. The outer circle has 8 small green spheres. Bromine forms covalent bonds because it has seven valence electrons, but neon has eight valence electrons and already fulfills the octet rule. Bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons. Neon forms covalent bonds because it can share its valence electrons, but bromine has seven valence electrons and can gain only one more electron. Neon forms covalent bonds because it has only two electron shells, but bromine has many electron shells and will lose electrons in order to fulfill the octet rule.
Physics
1 answer:
blondinia [14]2 years ago
6 0

Answer:

I believe the answer is Bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons.

Explanation:

valence electrons are the outermost shell, so when you go through keeping that in mind it helps you find the right answer

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Use the periodic table and your knowledge of isotopes to complete these statements.
klasskru [66]
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.

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7 0
2 years ago
Read 2 more answers
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
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LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

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  m is the mass of the car

  v is the velocity

  t is the time taken

  Force = \frac{1100 x 24}{2} = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

8 0
2 years ago
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