Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:
Equation 2, because K being more reactive, exchanges position with Pb in PbNO3.
Explanation:
Hello there!
In this case, according to the given reactions, it is possible to realize that according to the reactivity series, since K is is group 1A of alkali metals, we infer it is by far more reactive than magnesium, for that reason last two choices can be easily discarded. Now, considering equation 2, it would be necessary to complete it to figure out the correct option:

Whereas it can be seen that potassium exchanges position with Pb according to the double displacement reaction; therefore, the correct answer is "Equation 2, because K being more reactive, exchanges position with Pb in PbNO3".
Best regards!
Answer:
https://socratic.org/questions/diamond-has-a-density-of-3-52-g-ml-what-is-the-volume-in-cubic-centimeters-of-a-
<u>Answer:</u> The pH of the solution is 13.0
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of KOH = 716. mg = 0.716 g (Conversion factor: 1 g = 1000 mg)
Molar mass of KOH = 56 g/mol
Volume of solution = 130 mL
Putting values in above equation, we get:

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions
- To calculate hydroxide ion concentration of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
[tex[[OH^-]=0.098M[/tex]
Putting values in above equation, we get:

To calculate the pH of the solution, we use the equation:
pOH + pH = 14
So, pH = 14 - 1.00 = 13.0
Hence, the pH of the solution is 13.0