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Marrrta [24]
4 years ago
5

Which is most likely a covalent compound? A :LiF B :MgS C :NH3 D :CaCl2

Physics
2 answers:
Cerrena [4.2K]4 years ago
6 0
C: NH3

because they are both nonmetals, whereas the other options all have metals, Li, Mg, and Ca, these typically form ionic compounds.
laila [671]4 years ago
6 0

NH3 is the most likely covalent compound  

<h2>Further Explanation </h2>

Other Choices are:

  • LiF- Ionic compound  
  • MgS- ionic compound  
  • CaCl2- ionic compound  

<h3>Covalent compounds </h3>
  • These are compounds that are composed of non-metal atoms when two non-metals reacts to form a covalent bond. This occurs as a result of sharing of electrons between non-metal atoms involved.
<h3>Ionic compounds</h3>
  • Ionic compounds are formed when a non-metal reacts with a non-metal to form an ionic bond. It occurs as a result of transfer of electrons where the metal atoms loses electrons and the non-metal atom gains electrons forming a cation and an anion respectively.
<h3>Covalent bond   </h3>
  • This is a type of bond that is formed between non-metal atoms. It is formed as a result of sharing electrons between non-metal atoms involved.
  • When atoms involved contribute equal number of electrons to the bond formation, the type of bond is known as covalent bond. Like in this case, the atom with four outer most electrons may share its electrons with two oxygen atoms to form a compound with a double covalent bond.
  • However, when the shared electrons come from one atom then the bond is known as dative covalent bond.
<h3>Ionic bond   </h3>
  • This is a type of bond that occurs between metal ions and non-metal ions. Ionic bond occurs as a result of transfer of electrons from one metal atom to another non-metal atom.
  • After the transfer of electrons, metal atom loses electron to form a cation while the non-metal atom gains electrons to form an anion.

Other types of chemical bonds include;

  • Hydrogen bonds
  • Metallic bonds
  • Dipole-dipole interactions, etc.

Keywords: Covalent compound, covalent bond

<h3>Learn more about: </h3>
  • Covalent structures: brainly.com/question/10099699
  • Ionic bonding: brainly.com/question/5274289
  • Covalent bonding: brainly.com/question/5274289

Level: High school  

Subject: Chemistry  

Topic: structure and bonding  

Sub-topic: Covalent structures  

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A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6
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Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

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Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s. Use the in
-Dominant- [34]

Answer:

<em>The impulse is 2145 kg-m/s</em>

<em>The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.</em>

Explanation:

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Elapsed time of impact = 0.55 s

Impulse is the product of force and the time elapsed on impact

I = Ft

I is the impulse

F is force

t is time

For this case,

Impulse = 3900 x 0.55 = <em>2145 kg-m/s</em>

If the initial velocity was 2.95 m/s

and mass of car plus driver is 190 kg

neglecting friction, the initial momentum of the car is given as

P = mv1

where P is the momentum

m is the mass of the car and driver

v1 is the initial velocity of the car

initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s

We know that impulse is equal to the change of momentum, and

change of momentum is initial momentum minus final momentum.

The final momentum = mv2

where v2 is the final momentum of the car.

The problem translates into the equation below

I = mv1 - mv2

imputing values, we have

2145 = 560.5 - 190v2

solving, we have

2145 - 560.5 = -190v2

1584.5 = -190v2

v2 = -1584.5/190 = <em>-8.34 m/s</em>

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