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3 years ago

The density of gold is 19.3 g/cm3, and the density of iron pyrite is 5.02 g/cm3. What substance do you think you found?

Chemistry

1 answer:

Harrizon [31]3 years ago

4 0

Answer:

Step 1

1 of 2

V = m / D; V (nugget of gold) = 50 g / 5.0 g / cm 3 = 10 cm3; V (nugget of iron pyrite) = 50 g / 19 g / cm3 = 2.63 cm 3

Result

2 of 2

Due to these results you can conclude that nugget of gold which volume is 10 cm3 is bigger than volume of nugget of iron pyrite which is 2.63 cm3.

Explanation:

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Answer: The correct option is b.

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From the given information above, it can be stated that the correct option is b.

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What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

$E^o_{cell} =E^o_{cathode} -E^o_{anode}$

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

Learn more about electrode potential:

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Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V