That is false because aluminum melts at 2,470C
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Parasitism... basically a tick and a dog. The dog is the host, which is harmed by the tick.
Its <span>c.chromatography is the process of separating solutions on the basis of their boiling points </span>
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.
