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3 years ago

Va rog ! urgent! E pt chimie

Chemistry

1 answer:

n200080 [17]3 years ago

5 0

Answer:

An element

It stays shiny

Explanation:

Pure Gold is an element.

An element is a distinct substance that cannot be split-up into simpler substances. Such substances consists of only one kind of atom.

There are over a hundred elements known to date.

As an element, Gold is classified as a metal due to its very unique set of properties.

One of the indicator that gold does not react with oxygen is that it stays shiny. It does not give rusty look when exposed to air.

Substances that combines with oxygen have a rusty look or change appearance when expose to air. For example, iron.

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5. What could you do to convert from meters to centimeters? *

maria [59]

centi- is essentially 10^2 of one meter.

If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.

7 0

3 years ago

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A substance has a solubility of 350 ppm.how many grams of the substance are present in 1.0l of a saturated solution

sergeinik [125]

Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
S = 350 / 10^6
s = 3.5 x 10^-4
the multiply it to the total solution to calculate the amount of substance present
m = ( 3.5 x 10^-4 ) ( 1.01 )
m = 3.535 x 10^-4 g of the substance present

6 0

3 years ago

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Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

Describe why the overgrowth of algae in eutrophication is a bad thing.

Tanya [424]

Answer:

The overgrowth of algae consumes oxygen and blocks sunlight from underwater plants. When the algae eventually dies, the oxygen in the water is consumed. The lack of oxygen makes it impossible for aquatic life to survive

Explanation:

hope this helps have a good rest of your afternoon :) ❤

7 0

2 years ago

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When taking a measurement of temperature which tool should be used

Tcecarenko [31]

Answer:

A thermometer is the appropiate tool to use.

Explanation:

4 0

3 years ago