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soldier1979 [14.2K]
2 years ago
5

Which of the following is an oxidation-reduction reaction? so2(g) h2o(l) right arrow. h2so3(aq) caco3(s) right arrow. cao(s) co2

(g) ca(oh)2(s) h2co3(l) right arrow. caco3(aq) 2h2o(l) c6h12o6(s) 6o2(g) right arrow. 6co2(g) 6h2o(l)
Chemistry
2 answers:
Gennadij [26K]2 years ago
7 0

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

expeople1 [14]2 years ago
4 0

Answer:

correct option is (4).

Explanation:

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Explanation:

In solution H2SO3 produce H+ ions and SO3²⁻ ions. In the same way, NaOH produce Na⁺ and OH⁻ ions. The conductivity of a solution is directly proportional to the concentration of ions in a solution. During titration, you are adding more NaOH (That is, more Na⁺ and OH⁻ ions). But each moles of OH⁻ reacts with H⁺ ion producing H₂O. That means the moles of Na⁺ that you are adding = Moles of H⁺ are been consumed. The concentration of ions remains approximately constant. But, H⁺ ion conducts better than Na⁺ ion. That means before the equivalence point, conductivity is decreasing. But after the equivalence point you will add OH- ions in excess increasing ion concentration increasing the conductivity:

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3 years ago
Read 2 more answers
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Mariulka [41]

Answer:

T₂ = 182 K

Explanation:

Given that,

Initial pressure, P₁ = 120 kPa  

Initial temperature, T₁ = 0˚C = 273 K

We need to find the final temperature when the pressure is 80 kPa.

We know that, Gay Lussac's Formula is :

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{80\ kPa\times 273}{120\ kPa}\\\\T_2=182\ K

So, the new temperature is equal to 182 K.

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