pH value 1 represents a solution with the lowest OH⁻ion concentration.
<u>Explanation:</u>
pH is given by the expression as the negative logarithm to the base 10 of the concentration of hydrogen ions.
pH = -log₁₀[H⁺]
If the pH is lower than 7, pH < 7 then it is acidic
If the pH = 7, then it is neutral
If the pH > 7, then it is basic
If pH is 1 then the solution is showing mostly acidic character,which is least basic in its character.
So if the pH is 1, which is most acidic and least basic solution that is lowest OH⁻ ion concentration.
Answer:
it is actually b because i did this i picked b and got it right
Explanation:
CH2O2 formic acid I believe so
The correct answer really is B.
If you are directed to break that rule then you better be in a high level chemistry class. When I taught things like that I insisted that students just wait until the chemical permeated the fume cabinet and even then I was always very nervous.
Sometimes you have to know when to ignore a bad direction. If you are working with chlorine, for example, you should be especially careful. That stuff was used in WWI as part of a chemical warfare technique. Many men suffered grotesque deaths by breathing it in, particularly if they were in trenches. Chlorine is heavier than air. It sinks to the lowest level.
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
