Chemistry, 03.11.2020 14:00 bazsinghnagoke

Complete this nuclear reaction by selecting which particle would go in the blank.
19
Ne →
10
+ OF
19
9

Answers: 2

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Answer from: 19brendaddavenport

Missing particles: $^{0}_{1}e^{+}$ (a positron) and an electron neutrino $\nu_{\rm e}$ .

The nuclear equation would be:

${\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}$ .

Explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by .

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron $^{0}_{1}e^{+}$ and an electron neutrino $\nu_e$ would be released. On the other hand, in a beta-minus decay, a negatively-charged electron $\rm ^{0}_{1}e^{-}$ and an electron antineutrino $\overline{\nu}_e$ would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the $\rm Ne$ nucleus on the left-hand side is , meaning that the nucleus has a charge of +10 . On the other hand, the atomic number of the $\rm F$ nucleus on the right-hand side shows that this nucleus carries a charge of only .

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of . That rules out the possibility of the combination of one negatively-charged electron $\rm ^{0}_{1}e^{-}$ (with a charge of ) and an electron antineutrino $\overline{\nu}_e$ (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino $\nu_e$ , which carries no electric charge.)

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

the formula mass of a molecule or a compound is the sum of the atomic weights of the atoms in its empirical formula

(that's what i know about formula mass)

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Complete this nuclear reaction by selecting which particle would go in the blank.
19
Ne...