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\(4 \mathrm{~g}\) of copper was dissolved in concentrated nitric acid. The copper nitrate on strong heating gave \(5 \mathrm{~g}\) of its oxide. The equivalent weight of copper is a. 12 b. 16 c. 32 d. 34

Short Answer

Expert verified
The equivalent weight of copper is 32.

Step by step solution

01

Understand the Chemical Reaction

Copper dissolves in nitric acid to form copper nitrate, which upon heating decomposes into copper oxide. The reaction can be represented as: \( \text{Cu (s) + 4HNO}_3 (aq) \rightarrow \text{Cu(NO}_3)_2 (aq) + 2NO_2 (g) + 2H_2O (l) \) and \( \text{Cu(NO}_3)_2 \rightarrow \text{CuO} + 2NO_2 + 1/2 O_2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Copper Nitrate Decomposition
During the decomposition process, copper nitrate is broken down by heating, transforming into simpler substances. When we heat copper(II) nitrate \[ \text{Cu(NO}_3)_2 \],it undergoes decomposition:
  • Copper nitrate decomposes into copper oxide \( \text{CuO} \).
  • This reaction yields nitrogen dioxide \( \text{NO}_2 \) and oxygen \( \text{O}_2 \) as gaseous by-products.
Decomposition reactions like this one are important in chemistry because they reveal how complex substances can be simplified. Moreover, this particular reaction is endothermic, which means it absorbs heat, causing the copper nitrate to break down.
Understanding decomposition is crucial in studying equivalent weights in reactions. Equivalent weight is the mass of a substance that reacts with or supplies one mole of electrons in a redox reaction or one mole of hydrogen ions in an acid-base reaction.
Copper Oxide Formation
Copper oxide forms as a result of the decomposition of copper nitrate. When copper nitrate is heated, it shifts molecular arrangements, allowing for the stable formation of copper oxide. Copper(II) oxide, \( \text{CuO} \),is a black powder that results from this reaction.
  • Copper oxide is formed as a solid residue when copper nitrate is heated strongly.
  • This reaction is part of a broader category called thermal decomposition—substances breaking down due to heat exposure.
Copper oxide is not just the product of this decomposition, but also an important industrial compound. It is used in various applications, such as producing pigments, as a catalyst in chemical reactions, and sometimes in batteries.
Its formation is essential to accurately determining the equivalent weight of copper from the reaction.
Chemical Reactions in Nitric Acid
When copper is dissolved in concentrated nitric acid, a chemical reaction occurs that produces copper nitrate, nitrogen dioxide gas, and water. The reaction can be expressed as:\[\text{Cu (s) + 4HNO}_3 (aq) \rightarrow \text{Cu(NO}_3)_2 (aq) + 2NO_2 (g) + 2H_2O (l) \]
  • Concentrated nitric acid acts as a strong oxidizing agent.
  • The copper reacts with the nitric acid, resulting in the emission of brown nitrogen dioxide gas, which is a characteristic feature of this reaction.
This type of reaction is an example of a redox reaction—involving both reduction and oxidation processes. Copper is oxidized as it loses electrons and changes from a metal to a nitrate.
This foundational understanding of the reaction's mechanism is vital. It helps to calculate the equivalent weight in such reactions, defining the amount of copper needed to achieve specific residues.

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Most popular questions from this chapter

A \(6.977 \mathrm{~g}\) sample of a mixture was analysed for barium ion by adding a small excess of sulphuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulphate, which was collected by filtration, washed, dried and weighed. If \(0.4123 \mathrm{~g}\) of barium sulphate was obtained, what was the mass percentage of barium in the sample? a. \(6.952 \%\) b. \(3.476 \%\) c. \(4.376 \%\) d. \(1.738 \%\)

Hardness of water sample is \(300 \mathrm{ppm} \mathrm{CaCO}_{3}\). Hence its molarity is a. \(0.30 \mathrm{M}\) b. \(0.003 \mathrm{M}\) c. \(0.030 \mathrm{M}\) d. \(0.0013 \mathrm{M}\)

(A): In CO molecule 12 parts by mass of carbon combine with 16 parts by mass of oxygen and in \(\mathrm{CO}_{2}, 12\) parts by mass of carbon combine with 32 parts by mass of oxygen. (R): When two elements combine separately with a fixed mass or a third element, then the ratio of their masses in which they do so is either the same or whole number multiple of the ratio in which they combine with each other.

Number of water molecules in the drop of water, if \(1 \mathrm{ml}\) of water has 20 drops and 'A' is Avogadro's number, is a. \(0.5 \mathrm{~A}\) b. \(0.5 \mathrm{~A} / 18\) c. \(0.05 \mathrm{~A} / 18\) d. \(0.05 \mathrm{~A}\)

(A): In \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+\) \(4 \mathrm{H}_{2} \mathrm{O}, \mathrm{MnO}_{4}^{-}\)acts as oxidizing agent and \(\mathrm{Fe}^{2+}\) acts as reducing agent. (R): The reactions nvolving simultaneous loss or gain of electron among the reacting species are called oxidation reduction reactions.

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