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<em>Metals are an excellent conductor of electricity and heat because the atoms in the metals form a matrix through which outer electrons can move freely. Instead of orbiting their respective atoms, they form a sea of electrons that surround the positive nuclei of the interacting metal ions.</em>
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The reduction of alkyne to an alkene in the first step allows the best reagent to be chosen for each subsequent step.
Describe reagents.
A reagent is merely an essential component of a chemical reaction, it should be mentioned. It is an ingredient that speeds up the reaction.
With H2 and Lindlar's catalyst, an alkyne is reduced to alkene as the initial step in this process. Alkene will then be brominated to produce allyl bromide as the next step.
In this instance, the required allyl alcohol will be produced via the reaction of allyl bromide with NaOH.
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In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
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Answer:
24 mol Cu
General Formulas and Concepts:
<u>Chemistry</u>
Explanation:
<u>Step 1: Define</u>
RxN: 2Cu (s) + O₂ (g) → 2CuO (s)
Given: 12 moles O₂
<u>Step 2: Stoichiometry</u>
<u /> = 24 mol Cu
<u>Step 3: Check</u>
<em>We are given 2 sig figs.</em>
Our final answer is in 2 sig figs, so no need to round.