Chemistry Forum

What a big deal

4 C10H10BrNO3 + 47 O2 => 40 CO2+ 4 HBr+ 4 NO2+ 18 H2O

Sorry it is nonsense what you say. Even Chloroplatinic acid is used as an catalyst it is inorganic and not organic. If you don't know the difference then your expertise is poor.

By the way we dont like comercial Postings.

Convert mM in mg/l, use m = n* M
The density is not necessary, because its a solid material. You can only dissolve a powder in water. No volume.
Then calculate it for 10 ml.

Its ok , but there are no half molecules even mathematicaly correct

Times two

2 C2H6 + 7 O2 => 4 CO2 + 6 H2O

Yes of course it will work. But you have to know which one you will make. Citric acid has 3 Protons, these can be exchanged step by step by a metal, here Calcium, which is bivalent. So there will be CaHCit., Ca(H2Cit)2 and Ca3(Cit)2.

We can not give answers for pharmaceutical purpose.

(Hydroxy)-citric acid can be neutralized with Calcium carbonate to the citrate.

You cannot determine different Isotops of one element by titration. If you check for potassium then you will get all Isotopes together. The Isotopes are different by the amount of neutrons in the atom, this is not possible to figure out by chemical analysis. The chemical behaviour is driven by the a...

In this case convert all to KCl. pH 7.

Probably the determination of molarity of HCl is also wrong. If it was only the half instead 0.0098 M only 0.0049 M then it would more or less fit.

The solution p Value is sum of KOH and 1/2 K2CO3 or KHCO3 ( ml HCl consumed Phenolphthalein) m value is sum of KOH and K2CO3 ( ml HCl consumed Methylorange) n (KOH) = 2 p - m = 2*(KOH + 1/2 K2CO3) - (KOH + K2CO3) = KOH n ( K2CO3) = 2*(m-p) = 2*(KOH + K2CO3 - (KOH + 1/2 K2CO3)) = K2CO3 n KOH = 2 * 26...

Values for KOH and K2CO3 have to be exchangend.
See here an example with sodiumhydroxide and Carbonate

https://www.linkedin.com/pulse/analytic ... oda-kartal

You need about 355 ml solvent for the 0.7 g.

But read how to make it

What is aqua Regina it is 3 part HCl and one part HNO3 .

https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO3 is evaporated as NO2, later dissolve the product again in conc. HCl to get 355 ml

I thing to figure out what is aqua Regina and the calculati8n you have to do by yourself.

Calculate the moles of Gold you have then covert this to the volume of the acid to get 0,01 M

Dissolve in aqua Regina.

Yes something doesn't fit.
The titration values are wrong or the method to get molarity of HCl.

It was asked for potassium not for potassium carbonate or hydroxide.
The moles for K2CO3 are wrong

K2CO3 + 2 HCl => 2 KCl + H2O + CO2

For 1 K2CO3 2 HCl needed or 1 HCl correspond to 1/2 K2CO3

n (K2CO3) =0,0261 *0,09817/2 = 0,001281 mol