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ki77a [65]
3 years ago
5

What kind of electrical charges do protons,electrons, and neutrons have?​

Chemistry
1 answer:
Vinil7 [7]3 years ago
4 0

<em><u>Protons</u></em><em><u> = Positive Charge</u></em>

<em><u>Neutrons</u></em><em><u> = Neutral Charge/No Charge</u></em>

<em><u>Electrons</u></em><em><u> = Negative Charge</u></em>

<em>This one's simple: electrons have a negative charge, protons have a positive charge and neutrons — as the name implies — are neutral.</em>

<u><em>Protons</em></u>

<em>Elements are differentiated from each other by the number of protons within their nucleus. For example, carbon atoms have six protons in their nucleus. Atoms with seven protons are nitrogen atoms. The number of protons for each element is known as the atomic number and does not change in chemical reactions. In other words, the elements at the beginning of a reaction -- known as the reactants -- are the same elements at the end of a reaction -- known as the products.</em>

<em />

<em><u>Neutrons</u></em>

<em>Although elements have a specific number of protons, atoms of the same element may have different numbers of neutrons and are termed isotopes. For example, hydrogen has three isotopes, each with a single proton. Protium is an isotope of hydrogen with zero neutrons, deuterium has one neutron, and tritium has two neutrons. Although the number of neutrons may differ between isotopes, the isotopes all behave in a chemically similar manner.</em>

<em />

<u><em>Electrons</em></u>

<em>Electrons are not bound as tightly to the atom as protons and neutrons. This allows electrons to be lost, gained or even shared between atoms. Atoms that lose an electron become ions with a +1 charge, since there is now one more proton than electrons. Atoms that gain an electron have one more electron than protons and become a -1 ion. Chemical bonds that hold atoms together to form compounds result from these changes in the number and arrangement of electrons.</em>

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal
Sunny_sXe [5.5K]

Rydberg formula is given by:

\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )

where, R_{H} = Rydberg  constant = 1.0973731568508 \times 10^{7} per metre

\lambda = wavelength

n_{1} and n_{2} are the level of transitions.

Now, for n_{1}= 2 and n_{2}= 6

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )

= 1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.0278 )

= 1.0973731568508 \times 10^{7} \times 0.23

= 0.2523958\times 10^{7}

\lambda = \frac{1}{0.2523958\times 10^{7}}

= 3.9620\times 10^{-7} m

= 396.20\times 10^{-9} m

= 396.20 nm

Now, for n_{1}= 2 and n_{2}= 5

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.04 )

= 1.0973731568508 \times 10^{7} \times (0.21 )

= 0.230 \times  10^{7}

\lambda= \frac{1}{0.230 \times 10^{7}}

= 4.3478 \times 10^{-7} m

= 434.78\times 10^{-9} m

= 434.78 nm

Now, for n_{1}= 2 and n_{2}= 4

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.0625 )

= 1.0973731568508 \times 10^{7} \times (0.1875 )

= 0.20575 \times 10^{7}

\lambda= \frac{1}{0.20575 \times 10^{7}}

= 4.8602 \times 10^{-7} m

= 486.02 \times 10^{-9} m

= 486.02 nm

Now, for n_{1}= 2 and n_{2}= 3

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.12 )

=  1.0973731568508 \times 10^{7} \times (0.13 )

= 0.1426585\times 10^{7}

\lambda= \frac{1}{0.1426585\times 10^{7}}

= 7.0097 \times 10^{-7} m

= 700.97 \times 10^{-9} m

= 700.97 nm



5 0
2 years ago
Read 2 more answers
La configuración electrónica mas probable para el anión J -1 del elemento J con Z = 17 es:
Elanso [62]

Answer:

Explanation:

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2 years ago
Will NaBr, CO2, Ca3(PO4), or KZS raise a liquid's boiling point the most when
Vladimir79 [104]

Answer:

K_2S

Explanation:

According to the boiling point elevation law described by the equation \Delta T_b = iK_bb, the increase in boiling point is directly proportional to the van 't Hoff factor.

The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.

NaBr would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.

Ca_3(PO_4)_2 is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.

K_2S would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.

CO_2 is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.

Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.

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For this reaction: 4 Al + 3O2 = 2 Al2O3
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0.347 mols, working out shown on photo

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What is the mass of 7.50 moles of magnesium chloride, mgcl2?
Natali5045456 [20]
MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2
8 0
3 years ago
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