Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
Where, is the initial concentration = 1.50 mol/L
is the final concentration = 1/3 of initial concentration = = 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-
<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Answer:
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> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
With increasing temperature of the chemical reaction
16(2)/74.1=.431
.431x100= 43.1%