Explanation:
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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:
+4
Explanation:
In PbO2, oxygen exhibits an oxidation number of -2 (since it's not a peroxide or superoxide):
Let the oxidation number of Pb be x. Then, for the compound to be neutral, the oxidation numbers of all atoms should add up to zero.
⇒ x + (−2) + (−2) = 0
x = +4
So the oxidation no. of Pb is +4.
I hope this helps.
Answers:
Density = 0.8 g/mol.
Given data:
v = 25 ml
m = 20 g
δ = ?
Solution:
Formula for calculating density is given as,
Density = Mass / Volume
putting values
Density = 20.0 g / 25 ml
Density = 0.8 g/mol.
Answer:
the answer is A An atomic orbital can only hold a maximum of 2 electrons, each with opposite spins
Explanation:
