After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
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:
they will eat a lot of fish and make other animals go hungry
Explanation:
logic
Answer:
AuCl
Explanation:
Given parameters:
Mass of Gold = 2.6444g
Mass of Chlorine = 0.476g
Unknown:
Empirical formula = ?
Solution:
Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.
Elements Au Cl
Mass 2.6444 0.476
Molar mass 197 35.5
Number of moles 2.6444/197 0.476/35.5
0.013 0.013
Divide by the
smallest 0.013/0.013 0.013/0.013
1 1
The empirical formula of the compound is AuCl
