Answer:
Aye bro you from discovery too
Explanation:
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
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
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
![E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V](https://tex.z-dn.net/?f=E%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20logQ%5C%5CE%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20log%5Cfrac%7B%5BSn%5E%7B%2B4%7D%5D.%5BFe%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D.%5BFe%5E%7B3%2B%7D%20%5D%7D%20%5C%5CE%3D0.617V-%5Cfrac%7B0.05916%7D%7B2%7D%20log%5Cfrac%7B%280.13%29.%280.0037%29%7D%7B%280.0023%29.%280.11%29%7D%20%5C%5CE%3D0.609V)
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
We are using vinegar ( a polar compound) and oil (non polar compound) in the making of mayonnaise. The two will form emulsion when mixed together and will be separated as two distinct layers
Here chef uses egg to hold or homogenize the two immiscible substances together.
Thus egg is stabilizing the colloid formed between vinegar and oil.
Answer:
C
Explanation:
When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:
pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])
Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.
a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92
pH = 14 - pOH = 14 - 4.92 = 9.08
b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57
pH = 14 - pOH = 14 - 4.57 = 9.43
c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) = 8.95
pH = 14 - pOH = 14 - 8.95 = 5.05
d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) = 8.59
pH = 14 - pOH = 14 - = 5.41