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.

1. 1.732 mol L⁻¹
2. 3.0 mol L⁻¹
3. 1.02 × 10⁻⁴ mol L⁻¹
4. 3.4 × 10⁻⁵ mol L⁻¹

The decomposition of NH₃ on a platinum surface is a zero-order reaction. The rates of production of N₂ and H₂ will be respectively:
(given ; k = 2.5 × 10^–4 mol^–1 L s^–1 ) 

The rate equation of a reaction is expressed as, Rate = \(k(P_{CH_{3}OCH_{3}})^{\frac{3}{2}}\)

(Unit of rate = bar min^–1)

The units of the rate constant will be:
1. bar^1/2 min    
2. bar² min^–1   
3. bar^–1 min^–2  
4. bar^–1/2 min^–1

$1.$ $6.67$ $\times {10}^{-2}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ ${\mathrm{s}}^{-1}$
$2.$ $2.67$ $\times$ ${10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ ${\mathrm{s}}^{-1}$ $$
$3.$ $4.67$ $\times$ ${10}^{-3}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ ${\mathrm{s}}^{-1}$
$4.$ $4.27$ $\times$ ${10}^3$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ ${\mathrm{s}}^{-1}$

For a reaction, 2A + B → C + D, the following observations were recorded:

The rate law applicable to the above mentioned reaction would be:

1. Rate = k[A]²[B]³

2. Rate = k[A][B]²

3. Rate = k[A]²[B]  

4. Rate = k[A][B]  

During a nuclear explosion, one of the products is ⁹⁰Sr with a half-life of 28.1 years. If 1µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, the amount of ⁹⁰Sr that will remain after 10 years in the now grown up child would be -

(Given ,antilog(0.108)=1.28)

1. 0.227 µg 

2. 0.781 µg 

3. 7.81 µg 

4. 2.27 µg 

Select the correct option based on statements below:

Match the items of Column-I with items of Column-II:

Codes:

Codes: