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yaroslaw [1]
3 years ago
9

All of the noble gases, Group 18, have eight valence electrons in its outer shell (excluding helium which only has two). Which o

f these would represent the oxidation number of the noble gases such as xenon and argon?
Physics
2 answers:
ikadub [295]3 years ago
6 0

Answer:

Zero or +2

Explanation:

The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?

Their normal oxidation number is zero but some have been shown to be reactive.

Anna007 [38]3 years ago
5 0

Answer:

Explanation:

Since oxidation number is derived from the number of electron in the outermost shell of an element and further more this oxidation number is a function of how much electron is needed by an element to achieve a duplet ( He) or octet configuration ( complete outer most shell which is the attributes of group  18 elements or noble gases) and they don't readily react with any compound.. Since Xenon and argon are noble gases , they will have their outer most shell completely feel hence their oxidation number will be zero (0)

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What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/
user100 [1]

Answer:

answer below

Explanation:

Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v= 739/162

v= 4.56

4 0
3 years ago
8459299
sergeinik [125]

Answer:

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy

3 0
2 years ago
A net force of +15 N changes the momentum of an object by +100 kg-m/s. What is the time over which the force is applied? (please
sattari [20]
As momentum / time = force
so; time = 100÷15

so your answer is 6.7 !!
3 0
3 years ago
Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

8 0
3 years ago
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