AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Explanation:
Let the mass of isoamyl acetate be 100g.
Moles of Carbon = 60.58/12 = 5.048mol
Moles of Hydrogen = 7.07/1 = 7.07mol
Moles of Oxygen = 32.28/16 = 2.018mol
Mole Ratio of C : H : O
= 5.048 : 7.07 : 2.018
= 5 : 7 : 2.
Hence the empirical formula of isoamyl acetate is C5H7O2.
Has only 7 electrons so it want to bond with other elements so its very reactive and unstable. Hope this helps
Answer:
1.5 moles of Fe produced.
Explanation:
Given data:
Moles of FeO react = 1.50 mol
Moles of iron produced = ?
Solution:
Chemical equation:
FeO + CO → Fe + CO₂
Now we will compare the moles of ironoxide with iron.
FeO : Fe
1 : 1
1.5 : 1.5
Thus from 1.5 moles of FeO 1.5 moles of Fe are produced.
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of
(weak acid) is,

initially conc. c 0 0
At eqm.

First we have to calculate the concentration of value of dissociation constant
.
Formula used :

Now put all the given values in this formula ,we get the value of dissociation constant
.



By solving the terms, we get

No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9