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
Nothing. Many salad dressings are a mixture of sugar and vinegar.
Molar mass Na = 23.0 g/mol
1 mol ---- 23.0 g
n mol ---- 69 g
n = 69 / 23.0
n = 3.0 moles
1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?
3.0 * 6.02x10²³ / 1
= 1.806x10²⁴ molecules
hope this helps!
Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
Answer:
x = 4.17
y = 1.86
Explanation:
0.62 = log(x)
x = 10^0.62 = 4.17 ( to the nearest hundredth)
0.62 = ln(y)
y = e^0.62 = 1.86 (to the nearest hundredth)
