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4 years ago

Look at the diagram of an electrolytic cell below.

Chemistry

1 answer:

Ipatiy [6.2K]4 years ago

8 0

<em>The cathode is Y</em>

<h3><em>Further explanation</em></h3>

Electrolysis uses electrical energy to carry out redox reactions that are not spontaneous.

The ions in the solution flowing electrically will move towards to opposite charge of the electrode

The electrolysis material is an electrolyte which can be a solution or a melt.

In positive pole electrolysis cells - the anode is the site of the oxidation reaction, while the negative pole - the cathode is the reduction reaction site.

The result of the reaction in the anode is based on a substance that easily oxidized while the reaction in the cathode is based on a substance that easily reduced.

Electrons (electricity) enter an electrolysis cell through the negative pole (cathode)

The negative ion from the solution will move towards the positive electrode and release the electrons around the positive electrode (oxidation) and the electrons flow to the negative pole

Whereas around the negative electrode, there is electron binding and a reduction reaction occurs

So if we see the picture the cathode is Y

<h3><em>Learn more</em></h3><h3><em>reaction related to electrochemistry brainly.com/question/3461108</em></h3>

<h3><em>Answer details </em></h3>

Grade: Senior High School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: cathode, anode, oxidation, reduction, negative pole, electrode

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Answer:

757.5 kPa

Explanation:

Boyle's law states that p1 * v1 = p2 * v2

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3 years ago

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Fed [463]

Explanation:

(a) Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

= $1.20 \times (RT)^{-1}$

= $\frac{1.20}{0.0820 \times 1073}$

= 0.01316

(b) As the given reaction equation is as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

$p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c) Now, expression for $K_{c}$ for the given reaction equation is as follows.

$K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

= $\frac{x \times x}{(a - x)}$

= \frac{x^{2}}{(a - x)}[/tex]

where, a = initial conc. of $CaCO_{3}$

= $\frac{22.5}{100} \times 9.56$

= 0.023 M

0.0131 = $\frac{x^{2}}{0.023 - x}$

x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

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4 years ago

The solubility of a solute and solvent at a given temperature is 30.65 g/100 mL.

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Answer:

613.0 g

Explanation:

2 L = 2000 mL

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3 0

3 years ago

Read 2 more answers

Please Answer! 10 Points!

MA_775_DIABLO [31]

1 is b
hope this helps

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3 years ago

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When 16.9 g KOH is dissolved in 90.8 g of water in a coffee-cup calorimeter, the temperature rises from 18.5 °C to 34.27 °C. Wha

juin [17]

Answer:

-431.5 J/g

Explanation:

Mass of solution = Mass of solute + mass of solvent

Solute is KOH while solvent is water.

Mass of KOH = 16.9 g

Mass of water = 90.8 g

Mass of solution = 16.9 + 90.8

= 107.7 g

Change in temperature (Δt) = 34.27 - 18.5

= 16.2 °C

Heat required to raise the temperature of water is released by dissolving KOH.

Therefore,

Heat released by KOH = m × s× Δt

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Heat released by per g KOH = 7293 J/16.9 g

= 431.5 J/g

As heat is released therefore, enthalpy change would be negative.

Enthalpy change of KOH = -431.5 J/g

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