Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (
), in joules per gram-Kelvin, by the following model:
(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
, 
- Initial and final temperatures of water, in Kelvin.
If we know that 
, 
, 
and 
, then the change in entropy for the entire process is:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Answer:
Melting: the substance changes back from the solid to the liquid. Condensation: the substance changes from a gas to a liquid. Vaporization: the substance changes from a liquid to a gas. Sublimation: the substance changes directly from a solid to a gas without going through the liquid phase.
Explanation:
The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7.
Answer:
.11 mol
Explanation:
Convert mmHg to atms by dividing by 760. Then multiply 6.3 by the atms and divide by .08206*(273+28) to get mol
