Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Answer:
a. 59 m/atm
Explanation:
- To solve this problem, we must mention Henry's law.
- <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
- It can be expressed as: C = KP,
C is the concentration of the solution (C = 1.3 M).
P is the partial pressure of the gas above the solution (P = 0.022 atm).
K is the Henry's law constant (K = ??? M/atm),
∵ C = KP.
∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.
The answer will be 17.9 grams
1. Subscript is below
2. Coefficient large 2 indicates the number of moles
3. Atoms
1/2 O2 + H2 —> H2O
It’s the atoms that balance on each side
Notes that 1/2 is the coefficient and 2 is the subscript in H2 and H2O
Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7
