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Alexus [3.1K]
3 years ago
9

Fluorine-18 is a positron emitter used in pet scans. write a balanced nuclear equation for the reaction.

Chemistry
2 answers:
ohaa [14]3 years ago
8 0

Nuclear reaction: ¹⁸F → ¹⁸O + e⁺(positron) + ve(electron neutrino).

Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.

There are two types of beta decay: beta minus and beta plus.  

In beta plus decay, a proton is converted to a neutron, positron and an electron neutrino, so mass number does not change.  In this nuclear reaction, fluorine (atomic number 9) changes to oxygen (atomic number 8), because fluorine lost one proton.

In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino.

kow [346]3 years ago
5 0
<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it. Crude appearance. 18 18 0 F --> O + e 9 8 1 Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts. The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons. Followed by a right arrow to indicate the direction the reaction is going. Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons. Followed by a plus sign to indicate more. Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
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If the mass of 293nv is 293.10 amu and that of 295nv is 295.45 amu , what is the atomic weight of nv?
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Answer;

294.13 amu

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293.05+295.2=588.25

588.25/2= 294.13 amu

an amu of 294.13 using significant figures

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

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If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?
zavuch27 [327]
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