<span>B) The crystals did not phosphoresce within the drawer but did expose the film</span>
Answer:

Explanation:
By definition one <em>half-life</em> is the time to reduce the initial concentration to half.
For a <em>second order reaction </em>the rate law equations are:
![\dfrac{d[B]}{dt}=-k[B]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D-k%5BB%5D%5E2)
![\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BB%5D%7D%3D%5Cdfrac%7B1%7D%7B%5BB%5D_0%7D%2Bkt)
The <em>half-life</em> equation is:
![t_{1/2}=\dfrac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%7D)
Thus, substitute the<em> rate constant</em>
and the <em>half-life </em>time <em>224s</em> to find [A]₀:
![224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}](https://tex.z-dn.net/?f=224s%3D%5Cdfrac%7B1%7D%7B1.30%5Ctimes10%5E%7B-3%7DM%5E%7B-1%7D%5Ccdot%20s%5E%7B-1%7D%5BA%5D_0%7D)
![[A]_o=0.291M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.291M)
Answer: <span>Based on giveninformation, the compound is a
<u>Poly Unsaturated</u> fatty acid.
Explanation: Fatty acids are carboxylic acids with -COOH functional group and a chain of hydrocarbons.
Fatty Acids are classified as,
i) Saturated Fatty Acids: Those Fatty Acids in which there is no double bond between the carbon chain are called as saturated fatty acids.
Examples: Lauric Acid </span>CH₃(CH₂)₁₀COOH<span>
Myristic Acid </span>CH₃(CH₂)₁₂COOH<span>
Palmitic Acid </span>CH₃(CH₂)₁₄COOH
ii) Unsaturated Fatty Acids: Those Fatty Acids in which contain a single double bond (mono-unsaturated) or more than one double bond (poly-unsaturated) d between the carbon atoms in a chain are called as unsaturated fatty acids.
Examples: <span>
Linoleic acid</span>
<span>
Vaccenic acid </span><span>
Palmitoleic acid
The saturated fatty acid containing 26 carbon atoms is called as
Cerotic acid. While cerotic acid containing a double bond at position 3 and 9 has a IUPAC name
H</span>
exacosa-3,9-dienoic acid with following structure.
The reaction mechanism for an alpha,beta-unsaturated ketone to react with basic peroxide to form an epoxide is shown below with a general ketone. The basic hydroxide is used to deprotonate the peroxide molecule to create a strong HOO- nucleophile. The peroxide then attacks the beta-carbon of the alkene and this pushes the electrons up to the oxygen of the carbonyl. This is the first intermediate that is formed during this reaction.
After the intermediate is formed, the lone pair from the oxygen pushes back down to form the carbonyl once more and this breaks a carbon-carbon bond which attacks the oxygen of the peroxy group, ultimately substituting an -OH group and forming the final epoxide ketone product.
Answer:
A. neutron
Explanation:
Neutrons cannot be accelerated in an electric or magnetic field because they are neutral in charge.