To determine which compound is white in color, we need to examine the properties of each option. Let's evaluate them one by one:
From the above analysis, we can conclude that among the listed options, lead sulphate is distinctly known for its white coloration.
Therefore, the correct answer is lead sulphate.
Lead sulphate is white in color.
Ammonium sulphide is soluble.
Lead iodide is bright yellow.
Ammonium arsenomolybdate is yellow.
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: