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
Distilled water has the a neutral ph. Pure water always has a ph of 7.
The reaction for what was describe in the problem is:
N₂ + 3 O₂ --> 2 NO₃
The reactants involved are nitrogen and oxygen gas. From the word itself, oxygen is an oxidizing agent. <em>Therefore, this reaction is an oxidation reaction due to the presence of the oxidizing agent.</em>
Answer:
Methanol and butanol are alcohols. Alcohols have the same_______________ as alkanes and the __________ identifies the compound as an alcohol.
Explanation:
Alcohols belong to a group of organic compounds which contain -OH group as the functional group.
So alcohols have the same carbon -hydrogen bonds as alkanes and the -OH functional group identifies the compound as an alcohol.
