The area of the ring = π(R² - r²).
Given that this equals the area of the inner circle = πr², we have R² - r² = r², which gives R² = 2r², so R = r√2.
Therefore, the ratio R:r = √2:1.
The inscribed circle in the square has diameter 8 cm, so radius = 4 cm.
When a square is inscribed in this circle, its diagonal equals the diameter (8 cm).
If the diagonal of the smaller square is 8 cm, then using diagonal = side√2, we get side = 8/√2 = 4√2 cm.
Therefore, area = (4√2)² = 32 cm².
For a quadratic equation ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a.
Here, sum = -(k-3)/2 and product = (k-5)/2.
Given that sum = (1/2) × product, we have -(k-3)/2 = (1/2) × (k-5)/2.
Solving: -(k-3)/2 = (k-5)/4, which gives -2(k-3) = k-5, leading to -2k+6 = k-5, thus k = 11.
For second-order: t₁/₂ = 1/(k[A]₀); 100 = 1/(k × 0.5); k = 0.04 M⁻¹s⁻¹
1/[A] = 2 + 0.4(5) = 2 + 2 = 4; [A] = 1/4 = 0.25 M
Catalyst lowers Eₐ (forward and reverse) without changing ΔH, reaction enthalpy change
For elementary reactions, rate law exponents = stoichiometric coefficients. Rate = k[A]²[B]
If doubling concentration increases rate by 4 times (2²), reaction is second order
For first-order: k = 0.693/t₁/₂ = 0.693/30 = 0.0231 min⁻¹ ≈ 0.023 min⁻¹
Q₁₀ = k(T+10)/k(T) = (6 × 10⁻²)/(3 × 10⁻²) = 2. For typical reactions, Q₁₀ = 2-3