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Brums [2.3K]
3 years ago
13

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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Significant digits show the precision of measurements and calculations
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3 years ago
The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude
salantis [7]

\boxed{\sf \tau=rFsin\theta}

Put values

\\ \rm\hookrightarrow 40=2Fsin40

\\ \rm\hookrightarrow Fsin40=20

\\ \rm\hookrightarrow 0.64F=20

\\ \rm\hookrightarrow F=31.25N

8 0
2 years ago
The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func
Kitty [74]

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = 570 \times \dfrac{2\pi}{60}

              = 59.69 rad/s

1600 rpm =  = 570 \times \dfrac{2\pi}{60}

              = 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt    

167.6 = 59.7 + α x 7980

α  x 7980 = 107.9

α = 0.0135 rad/s²

8 0
3 years ago
Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a
nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

    p₀o = m₁ v₁ + m₂ v₂

    pf = 0

    m₁ v₁ + m₂ v₂ = 0

    m₁ / m₂ = -v₂ / v₁

    m₁ / m₂=  - (-6.2) / 4.7

     m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

    Xcm = 1/M    (m₁ x₁ + m₂ x₂)

We derive from time

   Vcm = 1/M   (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

    0 = 1/M   (m₁ v₁ + m₂v₂)

   m₁ v₁ + m₂v₂ = 0

    m₁ / m₂ = -v₂ / v₁

   m₁ / m₂ = 1.3

4 0
2 years ago
A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa
Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф =  angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the  circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0
2 years ago
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