Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Answer:
K⁺ (aq) + F⁻ (aq) + H⁺ (aq) + Cl⁻ (aq) → KCl (aq) + H⁺ (aq) + F⁻ (aq)
Explanation:
KF (aq) + HCl (aq) → KCl (aq) + HF (aq)
KF (aq) → K⁺ (aq) + F⁻ (aq)
HCl (aq) → H⁺ (aq) + Cl⁻ (aq)
KCl (aq) → K⁺ (aq) + Cl⁻ (aq)
HF (aq) → H⁺ (aq) + F⁻ (aq)
Answer:
Calcium Nitrate is made up of three different elements and contains a total of nine atoms. This compound's formula is Ca(NO3)2. There is one calcium atom, two nitrogen atoms, and there are six oxygen atoms in calcium nitrate
Atoms in covalent bonds do combine so as to be stable. As covalent bond consist non metals e.g O2 in this example each atom has vacance of 2 orbitals/ electrons so shairing electrons result their stability
Answer:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
The coefficients are 3, 1, 3, 1
Explanation:
From the question given above, the following data were:
Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
The above equation can be balanced as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:
AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
Thus, the equation is balanced.
The coefficients are 3, 1, 3, 1
