1. Option A. Temperature and Salinity
2. Option D. (I think)
3. Option D.
Hope your test goes well!
The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution.
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
<span>.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
</span><span>.8x = 14
</span><span>x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)
</span>
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
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° = 0.80 V - (-0.44 V) = 1.24 V
