Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
Hello!
The balanced equation for the
neutralization of KOH is the following:
HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)
To calculate the
volume of HCl required, we can apply the following equation:
So, the required volume of HCl is
541,54 mLHave a nice day!
Remember the order, <n, l, ml, and ms>.
n: energy level
l: subshell
ml: number of orbitals in the subshell.
ms: electron spin.
The first numbers (n) of both the electrons are the same, but the only difference is the second number (l). The first electron has l=2, indicating that the electron is in the "d" subshell. On the other hand, the second electron has l=1, indicating that the electron is in the "p" subshell.
*Remember*
l=0 (s) spherical shape
l=1 (p) peanut shaped
l=2 (d) clover
Answer:
(1) The sample should not be large, because a large sample would produce a higher and broader mp range.
(2) The rate of heating does not matter.
Explanation:
(1) The sample should not be large, because a large sample would produce a higher and broader mp range, because varying temperature range across the body will lead to inaccurate determination of melting point.
(2) In principle, the melting temperature is INDEPENDENT (not dependent) on the heating rate. so in other words, altering the heating rate does not affect the measure of melting point.
