The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ ⋅ mol − 1 51.00 kJ⋅mol−1 . Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ ⋅ mol − 1 Ea=136.00 kJ⋅mol−1 Catalyzed: A ⟶ B A⟶B E a = 85.00 k J ⋅ mol − 1 Ea=85.00 kJ⋅mol−1

The given question is incomplete.The complete question is:

The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ /mol Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ/mol Catalyzed: A ⟶ B A⟶B E a = 85.00 k J/mol. determine the factor by which tha catalysed reaction is faster than the uncatalysed reaction at 289.0 K if all other factors are equal.

The factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is

Explanation:

According to the Arrhenius equation,

The expression used with catalyst and without catalyst is,

where,

= rate of reaction with catalyst

= rate of reaction without catalyst

= activation energy with catalyst

= activation energy without catalyst

R = gas constant =

T = temperature =

Now put all the given values in this formula, we get

Therefore, the factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

2b2t

explanation:

2b2t