Answer:
4, 16,
Explanation:
SI2 is sulphur diiodide. Sulphur is in group sixteen (six valence electrons) while iodine is in group 17(seven valence electrons).
Since there are two iodine atoms and one sulphur atom, the molecule has twenty valence electrons. Out of these twenty valence electrons, only four are bonding electrons. The other sixteen electrons include the four nonbonding electrons found on sulphur and the twelve non bonding electrons found on the two iodine atoms having six nonbonding electrons each.
M1V1 = M2V2
M1 = 3.000 M
V1 = 0.8000 L
M2 = ?
V2 = 2.00 L
M2 = M1V1/V2 = (3.000 M)(0.8000 L)/(2.00 L) = 1.20 M
Answer:3
Explanation: if I was like a rated wave
Answer:
Explanation:
All three lighter boron trihalides, BX3 (X = F, Cl, Br), form stable adducts with common Lewis bases. Their relative Lewis acidities can be evaluated in terms of the relative exothermicities of the adduct-forming reaction. Such measurements have revealed the following sequence for the Lewis acidity: BF3 < BCl3 < BBr3 (in other words, BBr3 is the strongest Lewis acid).
This trend is commonly attributed to the degree of π-bonding in the planar boron trihalide that would be lost upon pyramidalization (the conversion of the trigonal planar geometry to a tetrahedral one) of the BX3 molecule, which follows this trend: BF3 > BCl3 > BBr3 (that is, BBr3 is the most easily pyramidalized). The criteria for evaluating the relative strength of π-bonding are not clear, however. One suggestion is that the F atom is small compared to the larger Cl and Br atoms, and the lone pair electron in the 2pzorbital of F is readily and easily donated, and overlaps with the empty 2pz orbital of boron. As a result, the [latex]\pi[/latex] donation of F is greater than that of Cl or Br. In an alternative explanation, the low Lewis acidity for BF3 is attributed to the relative weakness of the bond in the adducts F3B-L.
Answer:
I believe it's false because the atomic number is the number of protons in the nucleus of an atom.
