Answer:
HF
Explanation:
This concept can be understood from the knowledge of Intermolecular forces of attraction.
Intermolecular bonds are Van der Waals forces which are weak forces of attraction joining non-polar and polar molecules together. They exist in the form of London Dispersion Forces and Dipole-dipole attraction.
An example of Dipole-dipole attraction is the hydrogen bond which is a unique dipole-dipole attraction between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom such as fluorine, oxygen, or nitrogen).
Molecules that possess the characteristics of hydrogen bonding have a higher boiling point. In the given question, only HF undergo hydrogen bond due to the electronegative effect of the fluorine element.
F2 occurs as a weak London dispersion force and it occurs between non-polar molecules.
Consider the following equilibrium:
4NH₃ (g) + 3O₂ (g) ⇄ 2N₂ (g) + 6H₂O (g) + 1531 kJ
The given statement is True, because
According to Le Ch's Principle:
Systems that have attained the state of chemical equilibrium will tend to maintain their equilibrium state.
External factors such as the addition of products and reactants result in the disruption of the equilibrium state.
we expect the system to shift to the direction that offsets the change in concentration.
This results in the state of chemical equilibrium to be reestablished.
Hence, The statement is true,
- The addition of more ammonia (a reactant) would offset the state of equilibrium.
- To restore chemical equilibrium, the system must consume the excess reactants to form more products.
- A shift to favor the products side occurs.
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Answer:
c. chemistry
Explanation:
it uses science and math which are based on facts.
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
Oxidation
iron+oxygen happened
