Sunlight will be the answer
may be
(if you will provide a little more information about your question.....
it will be easy to give the answer)
Answer:
D) 1 iron(II), 2 chloride
Explanation:
Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;
Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2
The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.
Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.
Answer:
Depending on the thermometer, it may have the ability to go as high or low as melting, freezing or boiling point for water. Just make sure you know the boiling, melting and freezing points in Celsius, Fahrenheit and/or Kelvin and read your thermometer accordingly.
Explanation:
Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
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:
![Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_%7B3%7D%20%5D%20%2A%5BCl_%7B2%7D%20%5D%20%7D%7B%5BPCl_%7B5%7D%20%5D%7D)
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>
I believe the correct answer from the choices listed above is option A. The E+ of the cell would be 2.47 V. E of the cell is given by the formula E(cathode) - E(anode) . <span>So in this case the first half cell reaction is the reaction that takes place in the cathode and the second one takes place in the anode.</span>