The answer is B, sodium is an element.
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Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Answer:
2–methylpropene.
Explanation:
To successfully name the compound given in the question, we must observe the following:
1. Determine the functional group of the compound.
2. Locate the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent group attached and locate it's position by giving it the lowest possible count.
4. Combine the above to obtain the name of the compound.
Now, let us determine the name of the compound. This is illustrated below:
1. The functional group of the compound is the double bond i.e the compound is an alkene.
2. The longest continuous carbon chain is 3 i.e propene since it is an alkene.
3. The substituent group attached is methyl i.e CH3. In this case, we'll start counting from the side of the double bond being the functional group. Therefore, the methyl group i.e CH3 is at carbon 2.
4. Therefore, the name of the compound is:
2–methylpropene
