\(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\)
m moles 120 40 – 80 – 40
Heat liberated from reaction
= 40 × 10–3× 57 × 103 J …(1)
Heat gained by solution = \(mC\Delta T\)
Mass of solution \(m = V × d = 1000 × 1 = 1000 g\)
Heat gained by solution \(= 1000 × 4.2 × ΔT …(2)\)
From eq (1) and eq (2)
Heat liberated = Heat gained
\(40 × 10–3 × 57 × 103 = 1000 × 4.2 × ΔT\)
\(ΔT = 54 ×\) 10–2 \(°C\)
\(ΔT = 54°C\) (Rounded off to the nearest integer)
So, the answer is 54°C.
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
Gas | CO₂ | Ar | HCHO | CH₄ |
---|---|---|---|---|
\(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
Match List I with List II:
Choose the correct answer from the options given below:
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
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:
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: