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4 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

2 answers:

Akimi4 [234]4 years ago

5 0

Answer: C. It is always the same as it was before the collision.

Explanation:

The momentum, p, of an object is defined as the product of its mass, m, and its velocity, v: p = mv

The law of momentum conservation states that when there are no outside forces the total momentum of the system must be conserved.

Hence, when two objects collide and bounce apart, and no outside forces act on the system, the total momentum after the collision is always the same as it was before the collision.

This is a very important fact which permits you to deal with collision problems.

There are two types of collisions: inelastic collisions and ellastic collisions. While in elastic collisions you can use the conservation of mechanical energy to solve the problems, in inelastic collisions there is transformation of energy due to collision which implies that the mechanical energy is not conserved. Neverthelss, in both elastic and inelastic collisions, total momentum is conserved (it is the same after as it was before the collision), as long as no outside forces act on the system.

umka2103 [35]4 years ago

3 0

The answer would be C) it is always the same as it was before the collision

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Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol

77julia77 [94]

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Cl=1s^22s^22p^63s^23p^5$

$Cl^-=1s^22s^22p^63s^23p^6$

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

$Na=1s^22s^23p^63s^1$

$Na^+=1s^22s^23p^63s^0$

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

$Mg=1s^22s^23p^63s^2$

$Mg^{2+}=1s^22s^23p^63s^0$

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Ca= 1s^22s^22p^63s^23p^64s^2$

$Ca^{2+}=1s^22s^23p^6^23p^64s^0$

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

$K= 1s^22s^22p^63s^23p^64s^1$

$K^{+}=1s^22s^23p^6^23p^64s^0$

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5$

$Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2$

$Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0$

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

$F=1s^22s^22p^5$