Answer: The temperature of the gas reduced to 400K.
Explanation:
Stated that ; The pressure remains the same, that is initial and final pressure equals 1atm.
Applying Charles Law
Initial volume V1 = 1
Final volume V2 = 1/2 (halved)
Initial temperature T1 =800K
Final temperature T2 = ?
(1/800) = (1/2)/T2
T2 = 800/2
T= 400K
Therefore, when the volume is halved, the temperature reduced also to half ( 400K)
Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
It is a <u>mixture</u> of minerals mainly of corps that crystalizes of magma below earths surface.
Answer:
The activation energy is 164.02 kJ/mol
Explanation:
Log (k2/k1) = Ea/2.303R × [1/T1 - 1/T2]
k1 = 8.9×10^-4 s^-1
k2 = 9.83×10^-3 s^-1
R = 8.314 J/mol.K
T1 = 540 K
T2 = 578 K
Log (9.83×10^-3/8.9×10^-4) = Ea/2.303×8.314 × [1/540 - 1/578]
1.043 = 6.359×10^-6Ea
Ea = 1.043/6.359×10^-6 = 164020 J/mol = 164020/1000 = 164.02 kJ/mol
