Answer:
No, because Flourine can only form 1 bond, thus backbonding is not obtainable
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
15.Potassium oxide
16.Calcium chloride
17.Aluminium sulphide
18.CaS
Explanation:
15.K is the chemical symbol of Potassium and generally the name of the non-metal at the end of a formula has the suffix '-ide' and since O is oxygen, the name becomes Potassium oxide.
16. The same applies here. Ca is Calcium and Cl is Chlorine but since its the non-metal at the end, it ends in -ide. So Calcium chloride.
17.The same applies here too. Al is Aluminium and S is Sulphur so Aluminium sulphide.
18. Calcium's symbol is Ca and that of Sulphur is S and that gives the formula CaS.
I believe that is a plant called Wisteria.
Water would have a much lower boiling point much like its other hydrides and it would loss its ability to dissolve polar substances plus it couldn't form water columns so no more cohesion between water molecules
