Na + NaNO3 = Na2O + N2
4 Na + 2 NaNO3 = 6 Na2O + N2
6 Na on each side
2 N on each side
6 O on each side
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
It has: 2 atoms of sodium (Na) 1 atom of Carbon 3 atoms of Oxygen bound together with ionic and polar-covalent bonds
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.
