The reaction is shown below,
Step 1: Hydration of Alkene: In first step
Ethene is hydrated to Ethanol through
Hydroboration Reaction.
Step 2: Oxidation of Primary Alcohol: In this step partial oxidation of ethanol is carried out using mild Oxidizing agent <span>Pyridinium chlorochromate (
PCC) and Acetaldehyde is produced.
Step 3: Reduction of Acetaldehyde followed by Oxidation: In this step, first acetaldehyde is reduced to
secondary alcohol using
grignard reagent. After that the sec. alcohol is oxidized to
ketone by using oxidizing agent CrO</span>₃.
Answer : The new volume of the balloon is, 22.4 L
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
or,
where,
= initial volume of gas = 22.6 L
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above formula, we get the final volume of the gas.
Therefore, the new volume of the balloon is, 22.4 L
Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
