Answer:
Colourless
Explanation:
We know that Y^3+ has the electronic configuration of;
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (the 5s and 4d levels are empty).
According to the crystal field theory, the colour of complexes result from transitions between incompletely filled d orbitals.
As a result of this, complexes with empty or completely filled d orbitals are colourless. Thus, [Y(H2O)6]3 is colourless according to the Crystal Field Theory.
(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6
B. 8
Although it may seem that transition metals would have more than 8 valence electrons, they in fact do not.
Nitrogen triiodide<span> is the </span>inorganic compound<span> with the formula </span>NI3<span>. It is an extremely sensitive </span>contact explosive<span>: small quantities explode with a loud, sharp snap when touched even lightly, releasing a purple cloud of iodine vapor; it can even be detonated by </span>alpha radiation. NI3<span> has a complex structural chemistry that is difficult to study because of the instability of the derivatives.</span>
<span>The balanced chemical ionic equarion of MgS(aq)+CuCl2(aq) = CuS(s)+MgCl2(aq) would be as follows:
</span><span>Mg2+ + S2- + Cu 2+ + 2 Cl- ==> CuS + Mg2+ + 2 Cl-
Therefore, the net equation would be:
</span><span>Cu 2+ + S 2- ===> CuS
</span><span>
Hope this helps.
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