honestly, who knows. I just want to take a s.h.t.t right now.
Answer:
by increasing temperature
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:- 1.62 moles
Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.

from given data,
= 5.17 L,
= 1.05 moles
= 8.00 L,
= ?
Let's plug in the values in the formula:

On cross multiply:

= 1.62 moles
So, now the toy contains 1.62 moles of the air.