0.05 cm thick coating of silver is deposited on a plate of 0.05 m2 area. The number of silver atoms deposited on plate are _____ × 1023. (At mass Ag = 108,d = 7.9 g/cm³)
Step 1. Calculate the volume of silver coating:
\(Volume = 0.05 \times 0.05 \times 10000 = 25 \, \text{cm}^3\)
Step 2. Calculate the mass of silver deposited:
\(Mass\ of silver = 25 \times 7.9 \, \text{g}\)
Step 3. Calculate moles of silver atoms:
\(\text{Moles of silver } = \frac{25 \times 7.9}{108}\)
Step 4. Calculate the number of atoms:
\(\text{Number of atoms} = \frac{25 \times 7.9}{108} \times 6.023 \times 10^{23} = 11.01 \times 10^{23}\)
Let \( ABCD \) be a tetrahedron such that the edges \( AB \), \( AC \), and \( AD \) are mutually perpendicular. Let the areas of the triangles \( ABC \), \( ACD \), and \( ADB \) be 5, 6, and 7 square units respectively. Then the area (in square units) of the \( \triangle BCD \) is equal to:
If \( S \) and \( S' \) are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and \( P \) is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to: