Answer:
0.00011 JK.
The process does NOT violate the second law of thermodynamics
Explanation:
The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.
=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.
The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.
The change in entropy of the cold container = 1/295 = 0.00339 J/K.
Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.
Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.
The process does NOT violate the second law of thermodynamics.
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\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 of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol
![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH.
Exponentiate both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
<u>Answer:</u> The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.
<u>Explanation:</u>
There are three types of solution: acidic, basic and neutral
To determine the type of solution, we look at the pH values.
- The pH range of acidic solution is 0 to 6.9
- The pH range of basic solution is 7.1 to 14
- The pH of neutral solution is 7.
We are given:
Concentration of HI = 0.100 M
1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.100M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.100M)
Putting values in above equation, we get:
To calculate the pOH of the solution, we use the equation:
pH + pOH = 14
Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.
Answer:
A. Water particles barely move forward; they move in a circular pattern.
