Answer:
Ea= -175.45J
A= 3.5×10^14
k=3.64 ×10^14 s^2.
Explanation:
From
ln k= -(Ea/R) (1/T) + ln A
This is similar to the equation of a straight line:
y= mx + c
Where m= -(Ea/R)
c= ln A
y= ln k
a)
Therefore
21.10 3 104= -(Ea/8.314)
Ea=-( 21.10 3 104×8.314)
Ea= -175.45J
b) ln A= 33.5
A= e^33.5
A= 3.5×10^14
c)
k= Ae^-Ea/RT
k= 3.5×10^14 × e^ -(-175.45/8.314×531)
k = 3.64 ×10^14 s^2.
Answer:
These three factors are required for ionization potential or ionization energy.
Explanation:
Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.
Answer:
does your son ever talk to strangers?
Explanation:
Oxygen, fluorine and iodine are diatomic elements. Flourine is more reactive than the other two because it is the closest away to filling its outer layer of electrons and becoming stable like a noble gas.
