Is four electrons the answer you’re looking for?
Answer: Potassium Iodide, KI
Explanation:
Flame test colors:
Li+ = Crimson Red
Na+ = Bright Orange-Yellow
K+ = Lilac
Addition of nitric acid and silver nitrate (HNO3 and AgNO3),
Cl- = White precipitate
Br- = Creamy precipitate
I- = Yellow Precipitate
Hope this helps, brainliest would be appreciated :)
Answer:
The concentration of the CaBr2 solution is 96 µmol/L
Explanation:
<u>Step 1:</u> Data given
Moles of Calciumbromide (CaBr2) = 4.81 µmol
Volume of the flask = 50.0 mL = 0.05 L
<u>Step 2:</u> Calculate the concentration of Calciumbromide
Concentration CaBr2 = moles CaBr2 / volume
Concentration CaBr2 = 4.81 µmol / 0.05 L
Concentration CaBr2 = 96.2 µmol /L = 96.2 µM
The concentration of the CaBr2 solution is 96 µmol/L
The answer is c but it might be b it’s be
Answer:
sodium hexachloroplatinate(IV)- Na2[PtCl6]
dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br
pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2
Explanation:
The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.
The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.
