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Gelneren [198K]
1 year ago
11

Is a neutral particle a nucleus or electron cloud

Chemistry
1 answer:
Novay_Z [31]1 year ago
5 0

A neutral particle of an atom is located at the nucleus.

<h3>What are electron cloud?</h3>

Electrons are those negatively charged particles that are found revolving round an atom.

Electron cloud is defined as the model of an atom in which the atom consists of a small but massive nucleus surrounded by a cloud of rapidly moving electrons.

A neutral particle is a nucleus because the nucleus of an atom is made up of neutral particles found within the nucleus of an atom which doesn't contain any charges.

An atom is defined as the smallest and indivisible part of an element which can take part in a chemical reaction.

An atom is made up of nucleus which consists of the proton (which is positively charged) and the neutron (which bears no charges at all).

Therefore, a neutral particle can be said to be an electron cloud.

Learn more about atoms here:

brainly.com/question/6258301

#SPJ1

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How many molecules are in 4.2 miles of water?​
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What is the correct equilibrium constant expression for equation P2(g)
sertanlavr [38]

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k = [F2]² [PO]² / [P2] [F2O]²

Explanation:

In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:

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6 0
3 years ago
What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
Cerrena [4.2K]

Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

Thus, the de broglie wavelength is 2.037 \times 10^{-34} m.

                               

6 0
3 years ago
Read 2 more answers
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