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Angelina_Jolie [31]
3 years ago
14

Metals can be highly purified using electrolysis. Which equation best reflects the overall process of purifying copper?

Chemistry
2 answers:
WINSTONCH [101]3 years ago
7 0

Answer: The answer is D.  Cu(s)anode ---> Cu(s)cathode

Explanation: I just took the quiz

makvit [3.9K]3 years ago
3 0

Answer:

A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode

Explanation:

Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it.  The resulting changes at the anode and cathode are given by the equation:

cathode: Cu²⁺  + 2e⁻    ⇒     Cu

anode:    Cu     ⇒       Cu²⁺  + 2e⁻

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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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Based on the collision theory, which factors are likely to increase the rate of reaction? Select all that apply.
Alexus [3.1K]

Answer:

Option a

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