Answer:3480s⁻¹
Explanation:We can solve the following problem using the Arrhenius equation.
Arrhenius equation is given by:
![K=Aexp[-Ea/RT]](https://tex.z-dn.net/?f=K%3DAexp%5B-Ea%2FRT%5D)
A=Pre-exponential factor or frequency factor
Ea=Activation energy
R=Ideal gas constant
T=Temperature
K=Rate constant
From the Arrhenius equation we can see that the rate constant K is related with the activation energy and frequency factor.
In the question we are given with the following data:
Ea=42KJ/mol=42x 1000 J/mol
A=8.0×10¹ per second
T=298K
R=8.314J/K mol
when we substitute these given values in Arrhenius equation
![K=A{exp[-Ea \div RT]}\\K=8\ \times10^{10} s^{-1}{exp[-42000\div 8.314\times298]}\\K=8\ \times10^{10} s^{-1}{exp[-16.95]}\\K=4.35\times10^{^{-8}}\times8.0\times10^{^{10}}s^{-1}\\K=34.8\times10^{2}s^{-1}\\](https://tex.z-dn.net/?f=K%3DA%7Bexp%5B-Ea%20%5Cdiv%20RT%5D%7D%5C%5CK%3D8%5C%20%5Ctimes10%5E%7B10%7D%20s%5E%7B-1%7D%7Bexp%5B-42000%5Cdiv%208.314%5Ctimes298%5D%7D%5C%5CK%3D8%5C%20%5Ctimes10%5E%7B10%7D%20s%5E%7B-1%7D%7Bexp%5B-16.95%5D%7D%5C%5CK%3D4.35%5Ctimes10%5E%7B%5E%7B-8%7D%7D%5Ctimes8.0%5Ctimes10%5E%7B%5E%7B10%7D%7Ds%5E%7B-1%7D%5C%5CK%3D34.8%5Ctimes10%5E%7B2%7Ds%5E%7B-1%7D%5C%5C)
K=3480s⁻¹
The value of rate constant obtained is 3480s⁻¹.
Answer:
It's D
Explanation:
Sorry for the late answer. Since the squirrel ran constant speed for 10 seconds and it ran 20 meters the first point would be at (10,20) then since it stopped to rest for 15 seconds the next point would be (25,20) finally since it ran another 30 meters for 15 more seconds the next point would be at (40,50). Sorry if the answer isn't the best, i'm bad at explaining ;w;
Explanation:
An element just has one-type atoms/atom (e.g. O2). Meanwhile, a compound is a variety of atoms (e.g. H2O).
Answer:
i not anther stand your bla bla bla
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
