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
Answer:
<h3>The answer is 7.85 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

volume = final volume of water - initial volume of water
volume = 13.91 - 12 = 1.91 mL
We have

We have the final answer as
<h3>7.85 g/mL</h3>
Hope this helps you
Answer:
C. More NO2 and SO2 will form
Explanation:
Le Chatelier's Principle : It predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc
It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.
In given Reaction SO3 is introduced(increased) .
So equilibrium will shift in the direction where SO3 should be consumed(decreased)
Hence the equilibrium will go in backward direction , i.e

So more and more Of NO2 and SO2 will form
Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too