Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.
• ki
• fe2o3
• (nh4)3po4

KI is the chemical formula of potassium iodide. KI is a white solid salt under room temperature and pressure. Fe2O3 is the chemical formula of iron (III) oxide is a red-brown solid that exists under normal conditions. (NH4)3PO4 is called ammonium phosphate. This is a really unstable salt with no commercial value.

Answer :

The name of compound is, potassium iodide.

The name of compound is, iron (III) oxide.

The name of compound is, ammonium phosphate.

Explanation :

Ionic compound : It is a type of compound that is made up of the ions of the two different elements in which one element is a metal and another element is a non-metal.

The rules of ionic compounds is given by:

Positive ion is written first.The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(1) The given compound is an ionic compound. It is made up of two different ions which are potassium ion and iodide ion .

So, the name of is, potassium iodide.

(2) The given compound is an ionic compound. It is made up of two different ions which are iron ion and oxide ion .  In this an oxidation state of iron is (+3).

So, the name of is, iron (III) oxide.

(3) The given compound is an ionic compound. It is made up of two different ions which are ammonium ion and phosphate ion .

So, the name of is, ammonium phosphate.

  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.