In an Ellingham diagram, a negative slope indicates that the reaction becomes more favorable at higher temperatures, while a positive slope indicates that the reaction becomes less favorable.
Assertion A is correct: In an Ellingham diagram, the oxidation of carbon to carbon monoxide does indeed show a negative slope, indicating that the reaction becomes more favorable as temperature increases.
Reason R is incorrect: While CO is a reducing agent, it does not tend to decompose at higher temperatures; rather, the decomposition of CO to C and \( \text{O}_2 \) is unfavorable at high temperatures, as shown by its position on the Ellingham diagram.
Conclusion: The correct answer is 4.
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
Minerals are the naturally occurring, homogeneous inorganic solid substances. They are having a definite chemical composition and crystalline structure, hardness and color. For example, copper pyrite, calamine, etc.
Impurities in an ore are called gauge. The removal of a gauge from the ore is called concentration ore.
Several steps are involved in the extraction of pure metal from ores. Major steps are as follows –