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:
V₂ = 530.5 mL
Explanation:
Given data:
Initial temperature = 20.0°C
Final temperature = 40.0 °C
Final volume = 585 mL
Initial volume = ?
Solution:
Initial temperature = 20.0°C (20+273 = 293 K)
Final temperature = 40.0 °C (40+273 = 323 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 585 mL × 293 K / 323 K
V₂ = 171405 mL.K / 323 K
V₂ = 530.5 mL
Answer:
80L
Explanation:
V1/T1 = V2/T2
V2 = V1 T2/T1
T1 = 300K
V1 = 60L
T2 = 400K
V2 = ?
V2 = V1 T2/T1
V2 = (60L)(400K) / (300K)
V2 = 80L
Answer:
(a) 
(b) 
Explanation:
The reaction that is carried out by the enzyme catalase produces 
The reaction that is carried out by the enzyme catalase produces
