Answer:
M KIO3 = 1.254 mol/L
Explanation:
∴ w KIO3 = 553 g
∴ mm KIO3 = 214.001 g/mol
∴ volumen sln = 2.10 L
⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol
⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)
⇒ M KIO3 = 1.254 mol/L
Answer :
The Nernst equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Anode]}{[Cathode]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BAnode%5D%7D%7B%5BCathode%5D%7D)
where,
= standard cell potential
n = number of electrons in oxidation-reduction reaction
F = Faraday constant = 96500 C
R= gas constant = 8.314 J/Kmol
T = temperature
[Anode] = anodic ion concentration
[Cathode] = cathodic ion concentration
The compound that is formed is: MgI2
0.53313648 feet = 16.25cm
Correct answer is magnesium bromide. This is an ionic compound with metal forming a positive ion - K+ and halogen forming a negative ion - Br-. When group 7 element form ions they have a suffix -ide. Bromine is the element and when it forms a negative ion bromine is called bromide with the suffix.
Magnesium gives 2 electrons and bromine can take only 1 electron, therefore 2 bromine atoms are needed. Therefore magnesium forms ionic bonds with 2 bromine atoms.
The compound is called magnesium bromide
