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3 years ago

Where are the most reactive nonmetal elements found on the periodic table?

Physics

2 answers:

Sindrei [870]3 years ago

8 0

Answer;

At the top of group 17

Explanation;

Reactivity in chemistry deals with whether something will react with another substance. The reactivity of an element can be predicted based on its location on the periodic table.

The most reactive nonmetal is fluorine. Fluorine is a halogen, which is Group 17 on the periodic table, and the halogens are the most reactive nonmetals. This is because they all have one empty space in their valence electron shells. Of the halogens, fluorine's empty space is closest to the positively charged nucleus. This means it pulls harder on another atom's electrons to fill that empty space.

Llana [10]3 years ago

3 0

Answer: Option (a) is the correct answer.

Explanation:

Elements of group 17 are known as halogens whereas elements of group 18 are known as noble gases.

Noble gases have completely filled valence shell. Therefore, they are stable and does not react easily.

Whereas halogens have incompletely filled sub-shells, therefore, in order to gain stability they react readily with another atom.

Halogen atoms are also electronegative in nature. On moving down the group, the electronegativity decreases. Thus, the atom at the top of group 17, that is, fluorine is the most reactive non-metal because fluorine has high electronegativity so, it readily attracts an electron in order to completely fill its sub-shell.

Thus, we can conclude that at the top of group 17 most reactive nonmetal elements are found on the periodic table.

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How does the government involve in a child or young person’s upbringing in your<br> country?

dolphi86 [110]

The government are involved in a child or young person’s upbringing in my country by providing the right amenities and ensure the justice system is fair.

<h3>What is Upbringing?</h3>

This is defined as the way an individual treated at a young age by older adults such as parents.

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1 year ago

Explain how a star's absorption spectrum and the concept of red shift help support the Big Bang Theory.

marta [7]

Answer:

Red shift supports the big bang theory. ... The light from distant galaxies is red shifted (this tells us the galaxies are moving away from us) and the further away the galaxy the greater the red shift (this tells us that the more distant the galaxy the faster it is moving). Constellations look like they are moving because earth is rotating on it's axis.

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2 years ago

Location A receives

Vlad1618 [11]

Location A receives more rainfall than Location B due to the rain shadow effect.

<u>Explanation</u>:

Rain shadow effect is caused due to the presence of mountains.
A rain shadow area is an area of land that has been forced to become dry, devoid of any vegetation growth due to the blockage of precipitation by mountains. These rain shadow areas will have a dry climate.
The other side of the mountain would receive plenty of precipitation and therefore would be flourished with plant growth. These areas will have a cool and wet climate.
In this case, Location A is on the other side of the mountain and so receives more rainfall or precipitation. Meanwhile, Location B is on the rain shadow region and so receives less rainfall.

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3 years ago

The granulation pattern that astronomers have observed on the surface of the Sun tells us that: a. the Sun is a lot cooler on th

Norma-Jean [14]

Answer:

c. hot material must be rising from the Sun's hotter interior

Explanation:

Granulation is the grainy appearance of the solar photosphere produced by the top of the convection cells in the sun.

The grainy appearance are produced by granules on the photosphere of the sun and granules are caused by convection currents of plasma within the sun's convection zone.

The interior of these granules are brighter (and thus hotter) than the exterior of the granules which are darker.

<u>So, the granulation pattern that astronomers have observed on the surface of the Sun tells us that hot material must be rising from the Sun's hotter interior.</u>

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3 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago