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3 years ago

What is the word for the definition of an electron in the highest occupied energy level of an atom

1 answer:

6 0

Answer: The word for the definition of an electron in the highest occupied energy level of an atom is " valence electron".

Explanation:

A valence electron is an external shell electron associated with an atom in chemistry and physics that can participate in the creation of a chemical bond if the highest occupied energy level of an atom is not closed. All atoms in a single covalent bond add one valence electron to form a mutual pair.

The periodic table showcases the arrangement of valence electrons group and block wise like:

Alkali metals have n s 1 as external shell configuration like H, Li, Na, K etc.
Alkaline metals have n s 2 as external hell configuration like Be, Mg, Ca etc.
p-block comprises group 13 to 18 having general electronic configuration n s 2, n p 1–6.
d-block or transition metals have general electronic configuration (n-1) d 1–10, n s 1–2.

f-block or inner transition metals have general electronic configuration (n-2) f^1–14 (n-1) d^0-1 n s^2.

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Why is it warmer near the equator and cooler near the poles?

Phantasy [73]

It is because the equator is closer to the sun and because the sun's rays hit the surface of the Earth at a higher angle at the equator. The poles are colder because they don't get direct sunlight. The sun is always low on the horizon.

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4 years ago

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A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als

Norma-Jean [14]

Hi there!

Voltage in a series can be expressed by the following:

$V_T = V_1 + V_2$

In words, the total voltage is equal to the sum of the individual voltage drops in a SERIES circuit.

We can solve for the total voltage:

$V_T = 10 + 10 = \boxed{\text{ D. 20 V}}$

6 0

2 years ago

The whooping crane (Grus americana) is the tallest bird native to North America. It almost went extinct in the 1900s; in 1938, t

grin007 [14]

Answer:

The value is $a = 2.7183$

Explanation:

From the question we are told that

The relationship between the number of whooping cranes and the number of decades is $ln N = 2.85 + 0.039t$

The exponential relationship is $N = na^{kt}$

Now from the given equation we have that

$N = e^{2.85 + 0.039t}$

$N = 17.29 e^{0.039t}$

So comparing this equation obtained an the given exponential relationship we have that

$n = 17.29$

$k = 0.039$

$a = e = 2.7183$

$a = 2.7183$

7 0

3 years ago

A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.

Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

7 0

3 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago