Answer:
280.8 g
Explanation:
Definimos la reaccion:
2NaOH + FeSO₄ → Na₂SO₄ + Fe(OH)₂
Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.
Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol
2 moles de NaOH producen 1 mol de hidroxido ferroso
Entonces 6.25 moles producirán, la mitad (6.25 . 1) /2 = 3.125 moles
Convertimos los moles a masa:
3.125 mol . 89.85 g/mol = 280.8 g
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:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =

Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
= 6.02x10⁻⁶
First its ... higher energy of reactants , higher energy of products & you do the same with the lower ones than you do transition state
Answer:
In covalent bonding, the octet rule is important because sharing electrons gives both atoms a full valence shell. As a result, each atom can consider the shared electrons to be part of its own valence shell.
np :)