The following list contains some common radicals. using the charges on these ions and the idea of valence, predict the formulas for the following:
1. nitrate no3-1 phosphate po4-3
2. sulfate so4-2 acetate c2h3o2-1
3. ammonium nh4+ chromate cro4-2
4. carbonate co3-2 dichromate cr2o7-2
5. permanganate mno4-1 sulfide s-2
6. sodium carbonate
7. potassium permanganate

Answer : The formula for the compound sodium carbonate is

The formula for the compound potassium permanganate is

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non-metals.

(a) Sodium carbonate

In the compound, the sodium metal forms an ionic compound with polyatomic anion (carbonate) .

Here, sodium is having an oxidation state of +1 called as cation and carbonate is an anion with oxidation state of -2. The charges are not balanced. So, the charges are balanced by the cris-cross method. Thus, the compound formed will be,

Hence, the formula for the compound sodium carbonate is

(b) Potassium permanganate

In the compound, the potassium metal forms an ionic compound with polyatomic anion (permanganate) .

Here, potassium is having an oxidation state of +1 called as cation and permanganate is an anion with oxidation state of -1. The charges are balanced. Thus, the compound formed will be,

Hence, the formula for the compound potassium permanganate is

What is the gold assay

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.