Step 1: Initial Capacitance

The initial capacitance of the parallel plate air capacitor is given by:

Step 2: Equivalent Capacitance with Dielectric

When a dielectric of constant $K = 5$ fills half the distance ($\frac{d}{2}$), the system behaves as two capacitors connected in series:

• Air-filled portion ($C_1$):

  $$C_1 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2\varepsilon_0 A}{d} = 2C$$

• Dielectric-filled portion ($C_2$):

  $$C_2 = \frac{K \varepsilon_0 A}{\frac{d}{2}} = \frac{5 \cdot 2\varepsilon_0 A}{d} = 10C$$

Using the equivalent series capacitance formula:

Step 3: Percentage Increase

Molar mass of C$_{57}$H$_{110}$O$_6$: $57(12)+110(1)+6(16)=684+110+96=890\ \text{g mol}^{-1}$

Moles of C$_{57}$H$_{110}$O$_6 = \dfrac{445}{890} = 0.5\ \text{mol}$

From stoichiometry: 2 mol fat → 110 mol H$_2$O

So 0.5 mol fat → $\dfrac{110 \times 0.5}{2} = 27.5\ \text{mol H}_2\text{O}$

Mass of H$_2$O $= 27.5 \times 18 = \mathbf{495\ \text{g}}$

Tracing the folds:

• Fold 1 (diagonal): The paper is now a triangle. The folded corner = centre of original square.
• Fold 2 (triangle halved): The apex is still at the centre of the original square.

Punching at the folded corner (intersection of all folds = exact centre) means the punch goes through all 4 layers. But since the hole is at the centre, when unfolded it reveals 1 hole at the centre of the square — the layers were all stacked at that single point.

If the punch were at a non-fold point, you'd get 4 holes. But the apex/folded corner is a single geometric point — 1 hole appears.