The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>
The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R

K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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Explanation:
d. endothermic change as
heat is added to solid ice to change it to liquid water
If 0.5 L of solution contains 4 mol
then let 1 L of solution contain x mol
⇒ (0.5 L) x = (4 mol) (1 L)
x = (4 mol · L) ÷ (0.5 L)
x = 8 mol
Thus the molarity of the Sodium Chloride solution is 8 mol / L OR 8 mol/dm³.
<u>Answer:</u> The concentration of
comes out to be 0.16 M.
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the concentration of
comes out to be 0.1862 M.