Radius of the heap, r = \(\frac{10.5}{2}\) m = 5.25 m
Height of the heap, h = 3 m
Volume of the heap = \(\frac{1}{3}\pi\)r²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5.25 m × 5.25 m × 3 m
= 86.625 m³
Slant height,\(l = \sqrt{r² + h²}\)
\(= \sqrt{(5.25)² + (3)²}\)
\(= \sqrt{27.5625 + 9}\)
\(= \sqrt{36.5625}\)
= 6.046 m
∴ The area of the canvas required to cover the heap = \(\pi\)rl
= \(\frac{22}{7}\) × 5.25 m × 6.046 m
= 99.825 m²
Therefore, 99.825 m² canvas will be provided to protect the heap from rain.
When 3.0g of carbon is burnt in 8.00g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.0g of oxygen? Which law of chemical combination will govern your answer?