S sulfur
The element responsible for the yellow color of K2CrO4 is chromium. In this compound, chromium is in the +6 oxidation state which gives it a yellow color.
This is an impossible reaction. However if you turn it around and do AgNO3 plus K2CrO4 then it is equal to Ag2CrO4 plus 2KNO3.
The chemical formula for potassium chromate is K2CrO4.
K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. The oxidation number of oxygen is always -2 (except when chemically bound to fluorine, which is the only element higher in electronegativity/ionization energy than oxygen) K is an alkali metal, (in column 1A of the periodic table) and henceforth will almost always have an oxidation number of +1. With this information, we can deduct that... 2 K atoms will each have +1 as an oxidation number (totaling +2) 4 O atoms will each have -2 as an oxidation number (totaling -8) We set the equation to zero (because the compound has no charge) so. 2K+Cr+4O=0 or 2(+1)+Cr+4(-2)=0 or +2+cr-8=0 simple algebra from here. cr=+6 each, individually, would be, k=+1, cr=+6, o=-2. Generally, you'll only be looking to figure out the oxidation number of cr in this problem.
The dissociation equation for potassium chromate (K2CrO4) in water is: K2CrO4(s) -> 2K+(aq) + CrO4^2-(aq).
The element responsible for the yellow color of K2CrO4 is chromium. In this compound, chromium is in the +6 oxidation state which gives it a yellow color.
Chromate
Reaction of Potassium hydroxide or any potassium salt of a weak acid with chromic acid produces Potassium chromate. KOH + H2CrO4 ----> K2CrO4 + H2O K+ + H2CrO4 ----> K2CrO4 + H+
Oxidation of formic acid with K2CrO4 produces carbonic acid (HO COOH), but this molecule is very unstable, it tends to decompose in CO2 and H2O.
This is an impossible reaction. However if you turn it around and do AgNO3 plus K2CrO4 then it is equal to Ag2CrO4 plus 2KNO3.
The chemical formula for potassium chromate is K2CrO4.
K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. The oxidation number of oxygen is always -2 (except when chemically bound to fluorine, which is the only element higher in electronegativity/ionization energy than oxygen) K is an alkali metal, (in column 1A of the periodic table) and henceforth will almost always have an oxidation number of +1. With this information, we can deduct that... 2 K atoms will each have +1 as an oxidation number (totaling +2) 4 O atoms will each have -2 as an oxidation number (totaling -8) We set the equation to zero (because the compound has no charge) so. 2K+Cr+4O=0 or 2(+1)+Cr+4(-2)=0 or +2+cr-8=0 simple algebra from here. cr=+6 each, individually, would be, k=+1, cr=+6, o=-2. Generally, you'll only be looking to figure out the oxidation number of cr in this problem.
The dissociation equation for potassium chromate (K2CrO4) in water is: K2CrO4(s) -> 2K+(aq) + CrO4^2-(aq).
6. To determine this the CrO4 ion must be 2- as it is balanced by 2 potassium ions which can only have a +1 charge. The oxygen atoms are considered to be present as O2- so all four contribute 8-. As the net charge is only 2- then chromium must have charge (oxidation number or state) of 6+.
In K₂CrO₄, the oxidation number of potassium (K) is +1, and the oxidation number of oxygen (O) is -2. Since the compound is neutral, the oxidation number of chromium (Cr) can be calculated as follows: 2(+1) + Cr + 4(-2) = 0. Solving for chromium, the oxidation number of chromium in K₂CrO₄ is +6.
KCrO4 does not exist it is K2CrO4 and is called Potassium Chromate. Potassium dichromate is K2Cr2O7
To make a 0.25N K2CrO4 solution, you need to first calculate the molecular weight of K2CrO4 (potassium chromate). Then, determine the grams of K2CrO4 needed to make the desired volume of solution at a concentration of 0.25N. Dissolve this amount of K2CrO4 in the required volume of solvent, usually water, to make the final solution.