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3 years ago

Which of the following accurately describes properties of valence?

2 answers:

7 0

The answer is A), 'the smaller the number of electrons an atom has to borrow or to lend, the greater the activity of the atom'.

Troyanec [42]3 years ago

5 0

Answer: Option (A) is the correct answer.

Explanation:

Valence electrons are defined as the electrons present in the outermost shell of an atom.

Smaller is need for valence electrons to be borrowed or to lend more is the reactivity of an atom.

For example, atomic number of fluorine is 9 and its electronic distribution is 2, 7.

So, it just needs one more electron in order to gain stability. Therefore, it will be highly reactive in nature.

Thus, we can conclude that the smaller the number of electrons an atom has to borrow or to lend, the greater the activity of the atom, accurately describes properties of valence.

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Which situation is an example of increasing potential energy? Question 4 options: A. a cat jumping from a tree B. pulling a wago

jeka94

Pulling an wagon uphill I believe.

4 0

2 years ago

Read 2 more answers

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago

A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in

Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0

2 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

2 years ago

Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se

Nady [450]

Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω

7 0

1 year ago