Question

(A) The rate of the reaction 2NO + O2 → 2N02 at 25°C is 0.028 mol L-1 sec1 The experimental rate is given by r = [NO]? [02]. If the initial concentrations of the reactants are 02 = 0.040 mol L-1 and NO = 0.01 mol L-1, the rate constant of the reaction is​

Answer 1

Given :- $(9.64 \times 10 {}^{8} ) - (5.29 \times 10 {}^{6} )$

To Find :- Subtraction

Solution :-

$(9.64 \times 10 {}^{8} ) - (5.29 \times 10 {}^{6} )$

$= (9.64 - 5.29) \times (10 {}^{8} - 10 {}^{6} )$

$= 4.35 \times (10 {}^{8 - 6} )$

$= 4.35 \times (10 {}^{2} )$

$= 4.35 \times 100 = 435$

Hence, the subtraction is 435.

Hence, the final answer is 435