Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.
Answer:
yes the smoke from it scares them it also if, i remeber correctly, the smoke puts them to sleep
Explanation:
Answer:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
Explanation:
At a substitution reaction by SN1, the alkyl halide must lose its halide, and then an intermediary will be formed: a carbonium, which is an alkyl group with a positive charge in the carbon. The halide lost will be formed the halide ion, which is also an intermediary of the reaction.
The reactivity depends on the stability of the intermediaries (first of the carbonium, and second of the halide ion). As more bonded with carbons is the carbonium, more stable it is. The order of stability of the halides ions is from their electronegativity: as lower is it, as stable is the ion. The order is then: I⁻ > Br⁻ > Cl⁻ > F⁻.
2-bromo-2-methylpentane, 2-chloro-2-metylpentane, and 2-iodo-2-methylpentane, will form a 3-degree intermediary, so they will be more reactive than 3-chloropentane, which form a 2-degree intermediary. So, for the order of the stability of the halide ions, the order of reactivity is:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
Answer:
Explanation:
The chemical equation considered is:
- <em>Zn(s) + 2H⁺ (aq) → Zn²⁺ (aq) + H₂(g)</em>
That is a redox reaction because one both oxidation and reduction are happening.
The element or species that is being reduced is that whose oxidation number is reduced.
On the left side, hydrogen, H, has oxidation number +1, which is shown by the superscrit + to the right of H: H⁺.
On the right side, hydrogen appears in its elemental form: H₂. Thus, its oxidation number is 0.
That means that the hydrogen ion is being reduced, and the half reaction that represents this is:
Which means that two H⁺(aq) ions gain 2 electrons to produce a molecule of H₂(g). The reduction reactions happen due to the gain of electrons.
While the normal gas flame can
only produce a “operating” to “light blue” type of flame, the Bunsen burner can
at least yield three types of flame. Consequently, the following: <span><span />
Operating flame
– which is yellow/orange in color, near 300° C. </span>
<span><span>·
</span>
Blue flame –
can be imperceptible under normal lighting conditions, near 500° C. The typically
used laboratory type of flame.</span>
<span><span>·
</span>Roaring-blue
flame – forms a triangular shaped in the center of the flame normally light
blue in color and interestingly, it’s a sound-producing flame. Heat is near to
700° C. </span>
Imagine with this three kinds
of flame produced and a Bunsen burner creates compared to a simple normal gas
flame. In sense, the roaring-blue flame proves evident as to why Bunsen burner
is hotter hence, the amount of heat it makes (700°C) it makes.