Answer: H2 + 2Ag+ -> 2Ag + 2H+
Explanation:
H2 + Ag+ -> Ag + H+
Oxidation half reaction: H2 -> H+
H2 -> 2H+ + 2e-
Reduction half reaction: Ag+ -> Ag
Ag+ + e -> Ag
Balance the numbers of electrons in each half equations
H2 -> 2H+ + 2e- x1
Ag+ + e- -> Ag x2
H2 -> 2H+ + 2e-
2Ag+ + 2e- -> 2Ag
Combine both half equations
H2 + 2Ag+ + 2e- -> 2Ag + 2H+ + 2e-
Canceling out the equal number of electrons on both sides
H2 + 2Ag+ -> 2Ag + 2H+
Not sure but I would say because each orbital around the nucleus of the atom resembles a fuzzy cloud around the nucleus!
Answer:
The answer to your question is 4.14 g of O₃
Explanation:
Data
mass of nitrogen dioxide = 3.97 g
mass of ozone = x g
Reaction
NO₂ (g) + O₂ (g) ⇒ NO (g) + O₃ (g)
- Check that that reaction is balanced
Reactants Elements Products
1 N 1
4 O 4
This reaction is balanced
- Calculate the molecular mass of NO₂ and O₃
NO₂ = 14 + 32 = 46 g
O₃ = 48 g
- Calculate the mass of Ozone formed.
46 g of NO₂ ------------------ 48 g of O₃
3.97 gof NO₂ ---------------- x
x = (3.97 x 48 ) / 46
x = 190.56 / 46
x = 4.14 g of O₃
Answer:
Explanation:
The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"
The half-life of the element is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows: