For the reaction:

2N₂O₅ → 4NO₂ + O₂

At a given instant, the rate of reaction is 1.02 × 10⁻⁴ mol L⁻¹ s⁻¹ and the rate constant is 3.4 × 10⁻⁵ s⁻¹.
Calculate the concentration of N₂O₅ at that time.

The half-life period of $C^{14}$ is:
1.  500 years
2.  5000 years
3.  50 years
4.  5 × ${10}^4$ years

$\mathrm{In\; the\; reaction,\; 3}\mathrm{A}\to 2\mathrm{B\; the}$ rate of reaction $\frac{+\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}$ $$ is equal to:-

1. $-\frac{3}{2}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

2. $-\frac{2}{3}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

3. $-\frac{1}{3}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

4. $+2\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$