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Savatey [412]
2 years ago
5

An isotope contains 15 protons, 15 electrons, and 17 neutrons. What is the identity of the isotope?

Chemistry
2 answers:
mars1129 [50]2 years ago
7 0

Answer:

32P

Explanation:

Nuetrik [128]2 years ago
3 0

The isotope is phosphorus-32.  

It possesses 15 protons that make it phosphorus, 15 electrons, thus, the charges get balanced, and it is electrically neutral. It has 17 neutrons, so in combination, it has 32 protons/neutrons in the nucleus. Therefore, the name is phosphorus-32.  

The phosphorus-32 is a beta-emitter with a half-life of 14.3 days that is, used regularly in life-science projects, mainly to generate radiolabeled RNA and DNA probes, like for application in Southern Blots and Northern Blots.  


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Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

Formula used :

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

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Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

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a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

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True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

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therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

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brainly.com/question/17088180

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Explanation:

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