The reaction which shows oxidation and reduction simultaneously is C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l).
<h3>What are redox reactions?</h3>
Those reaction in which oxidation as well as reduction of substances takes place simultaneously will known as redox reactions.
- SO₂(g) + H₂O(l) → H₂SO₃(aq)
- CaCO₃(aq) → CaO(s) + CO₂(g)
- Ca(OH)₂(s) + H₂CO₃(l) CaCO₃(aq) + 2H₂O(l)
Above reaction are not the redox reactions as in these reaction oxidation and reduction simultaneously not takes place.
- C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)
In the above reaction reduction of oxygen takes place as its oxidation state changes from 0 to -2, and at the same time oxidation of carbon takes place as its oxidation state changes from 0 to +4.
Hence correct option is (4).
To know more about redox reactions, visit the below link:
brainly.com/question/7935462
Answer:
5.0x10⁻⁵ M
Explanation:
It seems the question is incomplete, however this is the data that has been found in a web search:
" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
NiCl₂ + 2AgNO₃ → 2AgCl + Ni(NO₃)₂
The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "
Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.
First we <u>calculate the moles of nickel chloride found in the 250 mL sample</u>:
- 3.6 mg AgCl ÷ 143.32 mg/mmol *
= 0.0126 mmol NiCl₂
Now we <u>divide the moles by the volume to calculate the molarity</u>:
- 0.0126 mmol / 250 mL = 5.0x10⁻⁵M
Answer:
all the statements are true of chemical changes
Answer:
chemical laboratory Pune Maharashtra India
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
