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3 years ago

True or False; the atomic number always equals the number of protons, neutrons and electrons found in an atom.

Chemistry

1 answer:

valina [46]3 years ago

8 0

"" ANSWER""

Protons are values of atomic numbers that do NOT change, that is, they are located inside the nucleus of the atom, whereas electrons are located in the electrosphere that can gain or lose electrons, the electron has a negative charge and the Proton positive, that is, if in any atom it wins an electron the atom will be negative and if the atom loses it will be positive because we already know that the number of protons does not change. To calculate the number of neutron we have to make the number of rounded atomic mass which becomes mass less the number of protons for example oxygen gas has 8 protons which is the atomic number and has atomic mass 15,999 which rounding up to 16 to find the number of neutrals we do 16-8 = 8 so now we know that oxygen has: 8 protons, 8 electrons and 8 neutrons but the values will not always be the same but the possibility of protons and neutrals having the same value is 75% by my count .

● ○ ● RULES ○ ● ○ ●

Atomic number = number of protons

Atomic mass ROUNDNESS = mass

Number of neutrons = number of protons - mass number

OBSERVATION- EXPLANATION FOR STUDENTS OF 9 YEARS BECAUSE IN THE HIGH SCHOOL UP SOME CHANGES CHANGE.

■■ GOOD STUDIES ■■

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Answer:

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Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

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$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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Learn more:

Expansion in gases brainly.com/question/9979757

#learnwithBrainly

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