Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:
You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
The concentrations are:
- [PCl₃]=
- [Cl₂]=
- [PCl₅]=
Replacing:
Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
<u>Answer:</u> The density of liquid is
<u>Explanation:</u>
We are given:
Mass of cylinder, = 65.1 g
Mass of liquid and cylinder combined, M = 120.5 g
Mass of liquid, =
To calculate density of a substance, we use the equation:
We are given:
Mass of liquid = 55.4 g
Volume of liquid = 49.3 mL = (Conversion factor: )
Putting values in above equation, we get:
Hence, the density of liquid is
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
dude! it's static!
Explanation:
a dry cell can't function and is therefore static!
<span>they exercise regularly, which improves the efficiency of their heart.
Hope this helps!
-Payshence xoxo</span>