Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
mass = 0.508 g, Volume = 0.175 L
Temperature = (25 + 273) K = 298 K, P = 1 atm
As per the ideal gas law, PV = nRT.
where, n = no. of moles =
Hence, putting all the given values into the ideal gas equation as follows.
PV =
1 atm \times 0.175 L =
= 71.02 g
As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of is 70.8 g/mol or 71 g/mol (approx).
Thus, we can conclude that the gas is most likely chlorine.
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).
The answer is [OH⁻] = 1 x 10⁻⁸.
To find OH⁻, divide the ionic product of water by [H₃O⁺] as :
<u>OH⁻ + H₃O⁺ = H₂O</u>
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- [OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
- [OH⁻] = 1 x 10⁻⁸