Step 1: Understanding the wave equation.
The given wave equation is:
\[ y = 3 \cos \pi (100t - x) \] This is a standard wave equation of the form: \[ y = A \cos(kx - \omega t) \] where \( k \) is the wave number and \( \omega \) is the angular frequency.
Step 2: Identifying the wavelength.
In the equation \( y = 3 \cos \pi (100t - x) \), comparing it with the standard form, we see that \( k = \pi \). The relationship between the wave number \( k \) and the wavelength \( \lambda \) is: \[ k = \frac{2\pi}{\lambda} \] Thus: \[ \pi = \frac{2\pi}{\lambda} \quad \Rightarrow \quad \lambda = 2 \, \text{cm} \] Thus, the correct answer is
(A) 2 cm.
At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.