need some one to check my answer .please ..chemistry?

Question

A KMnO4(aq) solution is to be standardized against As2O3(s) . A 0.1078g sample of As2O3 requires 22.15 ml of the KMnO4 (aq) for its titration . what is the molarity of the KMnO4(aq)?

5 AS2O3 + 4 MnO4 + 9 H2O + 12H ----> 10 H3AsO4 + 4 Mn
got M= 0.019680

is my answer right ???

Answer 1

http://answers.yahoo.com/question/index;_ylt=AjjxVMai2yElhXtEkVgZbjTsy6IX;_ylv=3?qid=20071118172237AAF6iep&show=7#profile-info-vXOyOzuGaa

Answer 2

Why might you think of there is any great difference between those 2 compounds. they have identically an identical layout and an identical molecular geometry. you have in elementary terms replaced C with Si and the hydrogen atoms have stayed an identical and the nitrogen stayed an identical. In the two cases N has 3 bonding pairs of electrons to 3 carbon atoms or 3 silicon atoms, and a lone pair. 3 bonds, and one million lone pair potential trigonal pyramidal. era. Why, in the 2nd undertaking, did you assert that N has in elementary terms 3 valence electrons? It nonetheless has 5. you do no longer propose B do you? in case you had (SiH3)3B, then it could be trigonal planar. ========= persist with up ========= i'd desire to respectfully disagree with Murty on 2 factors. First, VSEPR concept is genuinely clever for the two compounds. it incredibly is an exceedingly stable first-approximation for figuring out molecular geometry. Secondly, nitrogen is sp3 hybridized in the two compounds and is consequently trigonal pyramidal in the two compounds. there'll be little, if any, overlap between the lone pair electrons of nitrogen and the empty d-orbitals of Si. and whether there became into, it does not make the electrons "disappear" as Murty is suggesting. The electrons will nonetheless be around the nitrogen and that they're going to nonetheless be repelling the three bonding pairs of electrons, and the geometry of the electron pairs might nonetheless be tetrahedral with trigonal pyramidal molecular geometry. =========== greater persist with up =========== The paper that Timothy cites confirms the unlikelihood of the interplay of the p-orbital of N and the d-orbital of Si, and does propose that the (SiH3)3N is greater planar, making VSPER much less predictive for that reason. understand that the molecule has bond angles greater beneficial than the envisioned 109.5 tiers. this would be due greater to the dimensions of the SiH3 radical in assessment to the CH3 radical and the ensuing develop in repulsion. The paper potential that, because it says, "The planarity of (SiH3)3N is ascribed specially to electrostatic extremely than to d-orbital interplay." for that reason the consumer-friendly premise of VSEPR is valid to a point. What we see in the silicon based compound is that the repulsions of the SIH3radicals are super sufficient to conquer the repulsion of the lone pair, this forcing the SiH3 radicals farther aside, extremely than closer. this gives bond angles greater beneficial than 109.5 tiers and makes the molecule closer to being planar.