The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate.
Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.
The alkylation of diethyl malonate or a related ester of malonic acid at the carbon alpha (immediately next) to both carbonyl groups, followed by conversion to a substituted acetic acid, characterizes the chemical reaction known as the malonic ester synthesis.
As a result, it is evident from the structure of ethyl 2-methyl-4-pentenoate that ethyl and methyl bromides are the alkyl bromides employed.
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The solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L in water at 25°C.
At a specific solution temperature, a solid salt compound can entirely dissolve in pure water up to a predetermined molar solubility limit. The dissociation stoichiometry ensures that the molarities of the constituent ions are proportionate to one another. The saturable nature of the solution causes them to also coexist in a solubility equilibrium with the solid component. At this temperature, a solubility product constant Ksp is calculated using the solubility product of their molarity values.
Lead (II) fluoride has the following solubility equilibrium for its saturated solution:
⇄ 
![K_s_p = [Pb^2^+][F^-]^2](https://tex.z-dn.net/?f=K_s_p%20%3D%20%5BPb%5E2%5E%2B%5D%5BF%5E-%5D%5E2)
This compound dissociates in a 1:2 ratio of ions. For the compound dissolved in pure water, the Ksp is expressed in terms of the molar solubility "x" as:


Here,
× 
4.1 × 10⁻⁸ = 4 x³
x³ = 1.025 × 10⁻⁸
x³ = 10.25 × 10⁻⁹
x = 2.17 × 10⁻³ g/L
Therefore, the solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L.
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Answer:the answer is a because the chemicals do not change
Explanation:
Answer:
2-
Step-by-step explanation:
An O atom is neutral because it has the same number of protons as electrons.
O is in Group 16, so it <em>adds two electrons</em> to get a complete octet.
Since it has two extra electrons, the charge on an oxygen ion is 2-.
<span> 4 Al(s)+3 O</span>₂<span>(g) ------------> 2 Al</span>₂<span>O</span>₃<span>(s)
3 moles O</span>₂ -------------> 2 moles Al₂O₃
? moles O₂ --------------> 1.00 mole Al₂O3
O₂ = 1.00 * 3 / 2
O₂ = 3 / 2
= 1.5 moles of O₂
Molar mass O₂ = 32.0 g
1.5 * 32.0 = 48 g of O₂
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