Answer:
is usually structurally similar to the substrate.
Explanation:
Competitive inhibitors resemble normal substrate and binds to enzyme at the active site usually and prevents substrate from binding.
Active sites are main location for the substrate-enzyme binding. These sites involve weak as well as reversible bonds between the substrate and the enzyme. These inhibitors bind to the active sites and form weak and reversible bonds. Competitive inhibitors can be dissociated from active site by increasing concentration of the substrates. Substrates has to compete for active site and displace the bound competitive inhibitors.
<u>Hence, correct option is - is usually structurally similar to the substrate.</u>
Da na na na na na, nanana Na na na!
-Millie
Current, resistance and voltage.
Answer:
Enthalpy change = -44.12 kJ
Explanation:
<u>Given: </u>
ΔH°f(C2H6O(l)) = -277.69 kj/mol
ΔH°f(C2H4(g)) = 52.26 kj/mol
ΔH°f(H2O) = -285.83 kj/mol
<u>To determine:</u>
Enthalpy change for the formation of C2H6O
<u>Calculation:</u>
The given reaction is:
The enthalpy change for the reaction is given as;
where n(products) and n(reactants) are the moles of products and reactants
Substituting the appropriate values for n and ΔH°f:
![\Delta H = 1\Delta H^{0}f(C2H6O)-[1\Delta H^{0}f(C2H4)+1\Delta H^{0}f(H2O)]](https://tex.z-dn.net/?f=%5CDelta%20H%20%3D%201%5CDelta%20H%5E%7B0%7Df%28C2H6O%29-%5B1%5CDelta%20H%5E%7B0%7Df%28C2H4%29%2B1%5CDelta%20H%5E%7B0%7Df%28H2O%29%5D)
ΔH = -277.69-(52.26-285.83) = -44.12 KJ
