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Virty [35]
3 years ago
13

A radioactive nuclide is used to detect eye tumors. An atom of this radionuclide contains 15 protons, 15 electrons, and 17 neutr

ons. Which is symbol of this radionuclide?
Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0
Knowing that the number of protons of an element is equivalent to its atomic number, it can be determined from the periodic table of elements that this radionuclide is phosphorus. Since the number of its neutrons is 17, it can also be determined that its atomic mass is n + p = 15 + 17 = 32. This means the isotope is phosphorus-32.
lapo4ka [179]3 years ago
6 0
The answer is phosphorus-32.

We know that the atom contains <span>15 protons, 15 electrons, and 17 neutrons.
To conclude of which element this radionuclide is, we need to know an atomic number of the element. The atomic number of the element is the number of proton in the nucleus of the atom. If the number of protons is 15, then the atomic number is 15. The element with atomic number 15 is phosphorus (according to the periodic table).
To conclude which phosphorus isotope this radionuclide is we need to know the mass number of the element. The mass number of the element is the sum of the number of protons and number of neutrons in the nucleus of the atom. If the number of protons is 15 and the number of neutrons is 17, then the mass number is 17 + 15 = 32.
Thus, this radionuclide is phosphorus-32.</span>
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sineoko [7]

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<h3>What are nuclear equations?</h3>

The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.

Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.

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8 0
2 years ago
Two objects of the same mass are moving at the same velocity. You give
vladimir1956 [14]
I would imagine they’d be equal.
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3 0
3 years ago
What are the products obtained in the electrolysis of molten nai?
yanalaym [24]
Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


4 0
3 years ago
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