Chemistry, 08.04.2021 07:20 Rayzamartinez2757

In the following reaction, what stresses (changes) could we apply to the reaction in order to increase the production of H2O (g)? (hint: there are 4 correct choices) CO2 (g) + H2 (g) + heat ⇌ CO (g) + H2O (g)

CO (g) is added
CO (g) is removed
H2 (g) is added
H2 (g) is removed
CO2 (g) is added
CO2 (g) is removed
System (reaction) is heated
System (reaction) is cooled
Pressure is increased
Pressure is decreased

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Answer from: Quest

You know how the first one if filled up like the objects do the same but with the other categories

Answer from: Quest

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

Answer from: Quest

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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In the following reaction, what stresses (changes) could we apply to the reaction in order to increa...

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