Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

Chemistry

1 answer:

tamaranim1 [39]4 years ago

5 0

Answer:

This reaction is exothermic because the system shifted to the left on heating. Hope this helps! :D - RoseyCat :3

Explanation:

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Which option describes a phase change?

Leto [7]

I apologize of I'm wrong but its probably OA

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3 years ago

Plz help me with the electron dot structures! I’ll mark brainliest!

diamong [38]

The electron dot structure for each molecule are:
.. ..
a. H:O:O:H
.. ..
The partial negative charges are on the Oxygen atoms.

..
b. H:Cl:
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The partial negative charge is on the Chlorine atom.

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c. H:Br:
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The partial negative sign is on Bromine atom.

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d. H:O:H
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3 years ago

Complete this nuclear reaction by selecting which particle would go in the

kompoz [17]

Answer:

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 $1$ .

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 $10$ , 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 $+9$ .

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