The formation of Prussian blue is a qualitative test for the presence of ferric ions (Fe3+) in solution
The Prussian blue precipitate is formed when ferric ions (\( \text{Fe}^{3+} \)) react with hexacyanoferrate (\( [\text{Fe(CN)}_6]^{4-} \)) ions. The product of this reaction is \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \), which is an insoluble complex that imparts the characteristic deep blue color.
Step 1: Ferric chloride (\( \text{FeCl}_3 \)) dissociates to produce \( \text{Fe}^{3+} \) ions in solution.
Step 2: Potassium hexacyanoferrate (\( \text{K}_4[\text{Fe(CN)}_6] \)) dissociates to produce \( [\text{Fe(CN)}_6]^{4-} \) ions in solution.
Step 3: \( \text{Fe}^{3+} \) ions combine with \( [\text{Fe(CN)}_6]^{4-} \) to form the insoluble complex \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \), known as Prussian blue.
\( 4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 \downarrow \) (Prussian blue precipitate)
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
For hydrogen-like species, which of the following graphs provides the most appropriate representation of \( E \) vs \( Z \) plot for a constant \( n \)?
[E : Energy of the stationary state, Z : atomic number, n = principal quantum number]
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
The solutions can be classified into three types:
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types: