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3 years ago

Does potassium have a greater nuclear charge than Chlorine?

Physics

1 answer:

valentinak56 [21]3 years ago

6 0

Answer:

This is important because in the case of the potassium atom, the outermost electrons will be attracted by the nucleus more. In other words, K+ has bigger effective nuclear charge than Cl− , which translates to a bigger net positive charge felt by the outermost electrons.

Explanation:

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A temperature of 50.0 degrees Celsius is the same as a Kelvin temperature of approximately

Mila [183]

The answer is 323 K.

<u>Explanation:</u>

The Celsius temperature scale is equal to - 273.15.

One kelvin is equal to the change of thermodynamic temperature that results in a change of thermal energy kT by 1.380649 * 10^−23 J.

Convert degrees Celsius to kelvins with this simple formula:

kelvins = [degree C] + 273.15

The formula to convert Celsius to kelvins is degree C + 273.15.

50 degree celsius = 50 + 273.15

= 323.15 = 323 K.

4 0

4 years ago

Select all that apply. The resistance to an object being pulled over a surface is called _____. starting friction, sliding frict

jok3333 [9.3K]

That is kinetic friction.

8 0

3 years ago

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In order for a reaction to begin what is required

Andrew [12]

Activation energy?

are there multiple choices?

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3 years ago

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Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

3 years ago

Intersystem crossing is:

Semenov [28]

An intersystem crossing (ISC) is a non-radiative process that involves the transition between two electronic states with different spin multiplicity. That is, when an electron is excited in a molecule in a basal singlet state (either by absorption or radiation) into a state of greater energy, an excited singlet or triplet state can be obtained.

Therefore, ISC is understood as an a non radio active transition between states with different spin multiplicity.

Correct answer is C: a radiative transition between states with the same spin.

8 0

4 years ago