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VladimirAG [237]
3 years ago
8

1)Typically, atom gain or lose electrons to achieve (1 point)

Physics
2 answers:
solmaris [256]3 years ago
6 0
1)Typically, atom gain or lose electrons to achieve 
I believe the correct answer from the choices listed is option C, a stable electron configuration

2)The formation of an ionic bond involves

I believe the correct answer is A, transfer of electrons since it is electrons that is involved in a chemical reaction.

Flura [38]3 years ago
6 0

<em><u>Question 1:</u></em>

<u>Answer:</u>

C. Stable electronic configuration

<u>Explanation:</u>

According to Lewis, the atom will be stable when it has 8 electrons in its outermost energy level.

Therefore, atoms tend to undergo chemical reactions and form bonds by either losing, gaining or sharing electrons in order to fulfill the condition of having 8 electrons in outermost shell and become stable

<em><u>Question 2:</u></em>

<u>Answer:</u>

A. Transfer of electrons

<u>Explanation:</u>

Electrons are the part of the atom that are involved in bonding. Atoms (except for noble gases) tend to lose, gain or share electrons in order to have 8 electrons in outermost shell and achieve stability.

<u>Now, </u>

1- when an atom loses an electron or more to another atom that gains these electrons, <u>transfer of electrons</u> occur and <u>ionic bond</u> takes place

2- when atoms <u>share an electron or more</u>, a <u>covalent bond</u> occurs

<u>Based on the above</u>, transfer of electrons represents the formation of ionic bond

Hope this helps :)

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An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where
ankoles [38]

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s

6 0
3 years ago
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