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3 years ago

Which element does not have the same number of electrons in its outermost shell as the other elements in its group?(1 point)

Physics

2 answers:

Setler79 [48]3 years ago

4 0

Helium (He) does not have the same number of valence electrons as other elements in its group.

The periodic table is divided into groups with the last number of the group coinciding with the number of electrons that an element in the group has in its outermost or valence shell.

Helium is in group 18 which means that it should have the same number of valence electrons as :

Neon
Argon
Krypton
Xenon and,
Radon

Yet Helium only has 2 valence electrons. We can therefore conclusively say that Helium does not have the same number of valence electrons as other elements in its group.

More information is available at brainly.com/question/20944279.

artcher [175]3 years ago

4 0

Answer:

The answer is helium (He)

Explanation:

Because helium is in the group 18, and the other elements of group 18 have 8 electrons in it's outermost, while helium just have 2 electrons

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3 years ago

Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral

cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂) (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

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3 years ago

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