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4 years ago

How does iodine help in treating infections?

Chemistry

2 answers:

asambeis [7]4 years ago

6 0

Answer:

Iodine reduces thyroid hormone and can kill fungus, bacteria, and other microorganisms such as amoebas. A specific kind of iodine called potassium iodide is also used to treat (but not prevent) the effects of a radioactive accident.

Explanation:

lord [1]4 years ago

4 0

Answer:

i don't know

Explanation:

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Construct a flow chart to demonstrate how you could separate a mixture of 1,4- dichlorobenzene, 4-chlorobenzoic acid, and 4-chlo

notsponge [240]

Hey there!:

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4 years ago

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Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.

Nataly_w [17]

Answer:

$m_{sugar}=23.7g\ sugar$

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

$8oz*\frac{29.5735mL}{1oz} =236.6mL$

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

$m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar$

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3 years ago

a ray in the emission spectrum has a wavelength of 3.10x10^14 meters. given that the speed of light is 2.998x10^8m/. what is the

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The frequency of the ray with a wavelength of 3.10 × 10¹⁴ is 9.671 × 10⁻⁷.

I attached the working and the answer to the question below. I hope I was able to help.

Please note that C = speed of light, ν = frequency and λ= wavelength.

8 0

3 years ago

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 $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

Help pls this is due tomorrow

SSSSS [86.1K]

Answer:

10(HCl)

10H+10Cl

10+(35.453×10)

10+354.53

364.53

think it should be option a

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3 years ago

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How does iodine help in treating infections?​

How does iodine help in treating infections?