A nucleic acid whether DNA or RNA on complete hydrolysis, two purine bases, two pyrimidine bases, a pentose sugar and phosphoric acid.Nucleotides which are intermediate products in the hydrolysis contain
Correct Answer: Purine or pyrimidine base, a pentose sugar and ortho phosphoric acid
Elaborate Explanation:
Nucleic acids like DNA and RNA are long chains made of repeating units called nucleotides. When these nucleic acids are hydrolyzed (broken down using water), the final products are:
• 2 purine bases – Adenine (A) and Guanine (G)
• 2 pyrimidine bases – Cytosine (C) and Thymine (T) in DNA or Uracil (U) in RNA
• Pentose sugar – Deoxyribose (in DNA) or Ribose (in RNA)
• Ortho phosphoric acid – Also known as phosphoric acid (H₃PO₄)
During partial or intermediate hydrolysis, we get nucleotides, which are the building blocks of nucleic acids. Each nucleotide is made of:
• A nitrogenous base (purine or pyrimidine)
• A pentose sugar
• An ortho phosphoric acid group
Therefore, the intermediate product – a **nucleotide** – contains all three parts:
Purine or pyrimidine base + Pentose sugar + Ortho phosphoric acid
This is what makes a nucleotide different from a nucleoside, which has only the base and sugar (without phosphate).
Nucleotides are the building blocks of nucleic acids (DNA and RNA) and consist of three main components: a nitrogenous base (either a purine or pyrimidine base), a pentose sugar (ribose in RNA or deoxyribose in DNA), and a phosphate group (orthophosphoric acid). When nucleic acids undergo complete hydrolysis, these components are released as individual nucleotides.
Therefore, the correct answer is (C) a purine or pyrimidine base, a pentose sugar, and ortho phosphoric acid.
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:
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