Answer:
120.575 kJ is the activation energy for the souring process.
Explanation:
The formula for an activation energy is given as:
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get:l
![\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7Bk%7D%7B40k%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B298K%7D-%5Cfrac%7B1%7D%7B277%20K%7D%5D)
120.575 kJ is the activation energy for the souring process.
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Where is the word problems? Like the problems that you have to put it in.
Explanation:
