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natta225 [31]
3 years ago
10

Based on its location on the periodic table which element would be most likely to form a negative ion

Physics
2 answers:
Vinvika [58]3 years ago
8 0

Answer:

This question is incomplete but the completed question is below

Based on its location on the periodic table, which element would be most likely to form a negative ion?

A. Aluminum (Al)

B. Copper (Cu)

C. Sodium (Na)

D. Fluorine (F)

The correct option is D

Explanation:

An atom will become negatively charged when it accepts/gains electron(s). Atoms lose/gain electrons in order to achieve there octet configuration. In the periodic table, metals donate/lose their valence electron(s) to become positively charged (cation) while non-metals accept/gain electron(s) to become negatively charged (anion).

From the explanation above, it can be deduced that aluminium (Al) will lose the three valence electrons in it's outermost shell to become Al³⁺, copper (Cu) will lose two valence electrons to become Cu²⁺ while sodium (Na) will lose the only electron it has on its outermost shell to become Na⁺.

Fluorine (F), a non-metal, however gains an electron to become a negatively charged ion (F⁻).

blagie [28]3 years ago
6 0

Answer:

Since Fluorine is very electronegative, it can easily absorb the electrons of other elements. Since it sucks up electron, this gives Fluorine an excess electron thus making it a negative ion F-.

Explanation:

because it is

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A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) in
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Answer:

A) r = 0.03 m

B) r = 0.0533 m

C) B_max = 0.00003 T

Explanation:

Formula for magnetic field inside the capacitor when it is parallel to the length element is;

B_in = (μ_o•I•r/(2πR²)

Formula for maximum magnetic field is;

B_max = (μ_o•I/(2πR)

Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

A) Magnetic field inside the capacitor is gotten from our first equation above;

B_in = (μ_o•I•r/R²)

Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.

Thus;

B_in = 0.75B_max

(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))

μ_o•I and 2πR will cancel out to give;

r/R = 0.75

r = 0.75R

We are given R = 40 mm = 0.04 m

r = 0.75 × 0.04

r = 0.03 m

B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:

B_out = 0.75B_max

(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))

μ_o•I and 2π will cancel out to give;

1/r = 0.75/R

r = R/0.75

r = 0.04/0.75

r = 0.0533 m

C) B_max = μ_o•I/(2πR)

μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A

Thus;

B_max = (4π × 10^(-7) × 6)/(2π × 0.04)

B_max = 0.00003 T

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