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3 years ago

What trend do you note for the atomic radii of the third period?

Chemistry

2 answers:

Natasha_Volkova [10]3 years ago

6 0

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As you move across the period, the atomic radii decreases.

This is because of an increase in the number of protons (and electrons) and therefore, an increase in the nuclear charge.

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Law Incorporation [45]3 years ago

5 0

Answer:

The atomic radii decreases from left (sodium) to right (Argon).

Explanation:

The number of positive protons in the nucleus increases so the pull on the electrons increases and this decreases the radius.

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What is the mass of an object that requires a force of 25 N to accelerate at 5 m/s/s?

AfilCa [17]

The mass is 6 Kilograms

Explanation:

Force
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3 years ago

Generally, what happens to the rate of dissolution for a solid solute when you change from stirring a solution to not stirring i

Makovka662 [10]

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3 years ago

Read 2 more answers

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = E x r²

q = 1250 N/C x 0.205m²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = 5.84 x 10⁻⁹ C

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0

3 years ago

Idenify each statement below as either oxidation or reduction.

Alenkasestr [34]

Answer:

1 = oxidation

2 = reduction

Explanation:

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Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

2I- ----> I₂+ 2e⁻

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

F + e⁻ ----> F⁻

Consider the following reactions.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

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3 years ago

When energy is transferred from the air to the water, what happens to most of the energy?

Norma-Jean [14]

Most of the energy travels trough the water

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3 years ago