Choose the correct sets with respective observations:
(1) \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) → Blue precipitate
(2) \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) → Blue precipitate
(3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) → \( \frac{1}{2}K_4\text{Fe(CN)}_6 \)
(4) \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) → 6KC1
In the light of the above options, choose the correct set:
\( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) → 6KC1
To solve the given problem, we need to analyze each chemical reaction and identify which set of reactions is correct. Here are the options:
(1) \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) \(\rightarrow\) Blue precipitate
(2) \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) \(\rightarrow\) Blue precipitate
(3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) \(\rightarrow\) \( \frac{1}{2}K_4\text{Fe(CN)}_6 \)
(4) \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) \(\rightarrow\) 6KC1
Consideration of known inorganic chemistry reactions indicates that (3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) \(\rightarrow\) \( \frac{1}{2}K_4\text{Fe(CN)}_6 \) is the valid reaction.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: