Answer:
I never did this but i think its d 37.5٪ lmk if i got it right pls and srry if i didnt
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
For changing potential energy; When you drop a book, gravitational potential energy is transformed into kinetic energy. Your car transforms the chemical potential energy stored in gasoline into the kinetic energy of the car's motion.
For Kinetic Energy; kinetic energy can be converted into electrical energy by a generator or into thermal energy by the brakes on a car.
Answer:
VH2SO4 = 145.3 mL
Explanation:
Mw BaO2 = 169.33 g/mol
⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2
⇒according to the reaction:
mol BaO2 = mol H2SO4 = 0.545 mol
⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )
⇒V H2SO4 = 0.1453 L (145.3 mL)
Answer:
47.9 g of ethanol
Explanation:
Combustion is a chemical reaction in which a substance reacts with oxygen to produce heat and light. Combustion reactions have been very useful as a source of energy. Ethanol is now burnt for energy purposes as a fuel. Ethanol has even been proposed as a possible alternative to fossil fuels.
Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy
x moles of ethanol releases 1418 kJ/mol.
x= 1 × 1418 kJ/mol/ 1367 kJ/mol
x= 1.04 moles of ethanol.
Mass of ethanol = number of moles × molar mass
Molar mass of ethanol = 46.07 g/mol
Mass of ethanol = 1.04 moles × 46.07 g/mol
Mass of ethanol= 47.9 g of ethanol
