Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:
So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g
Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol.
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.
