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:
atomic mass of X is 48.0 amu
Explanation:
Let y be the atomic mass of X
Molar mass of O_2 is = 2×16 = 32 g / mol
X + O2 -----> XO_2
According to the equation ,
y g of X reacts with 32 g of O_2
24 g of X reacts with Z g of O_2
Z = ( 32×24) / y
But given that 24.0 g of X exactly reacts with 16.0 g of O_2
So Z = 16.0
⇒ (32×24) / y = 16.0
⇒ y = (32×24) / 16
y= 48.0
So atomic mass of X is 48.0 amu
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
Answer:
See pictures!
Explanation:
Here is a drawing of butane (C4H10); labeled #1. An isomer of butane will also have the chemical formula C4H10 but is depicted different geometrically. I labeled this one #2.