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oksano4ka [1.4K]

3 years ago

9

An oil tanker spills a large amount of oil near the ocean shoreline. Which application of chemistry would best solve this proble

m?

1 answer:

DochEvi [55]3 years ago

3 0

I have heard they can use hair is stocking or nets to absorb the oil out of the water

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A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?

olga nikolaevna [1]

The volume that will be occupied at 735 torr and 57 c is 23.12 L

<u><em>calculation</em></u>

<u><em> </em></u> At STP temperature=273 k and pressure=760 torr

<u><em> </em></u>by use of combined gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

T1= 273 K

V1= 18.5 L

P2= 735 torr

T2= 57+273= 330 K

V2=?

by making V2 the formula of subject

V2= T2P1V1/P2T1

V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L

5 0

3 years ago

The burning of coal, a fossil fuel, produces water and carbon dioxide. How is this similar to cellular respiration?

Mrac [35]

Answer:

I think the answer is C but you might need a second opinion on this answer

6 0

3 years ago

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What element represented by the bohr model below?

Svet_ta [14]

Answer:

B. flourine

Flourine is the 9th element in the periodic table

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]

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

<u>Explanation:</u>

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:

<u>Case 1:</u>

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 $E^o_{cell}$

<u>Case 2:</u>

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 $E^o_{cell}$

<u>Case 3:</u>

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

Calculate the number of moles equivalent to 12.7 gram of iodine molecule

Alexus [3.1K]

$\LARGE{ \boxed{ \purple{ \rm{Answer}}}}$

☃️ Chemical formulae ➝ $\sf{I_2}$

How to find?

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

$\boxed{ \sf{No. \: of \: moles = \frac{Given \: weight}{Molecular \: weight} }}$

Solution:

❍ Molecular weight of $\sf{I_2}$

= 2 × 126.90

= 253.80

= 254 (approx.)

❍ Given weight: 12.7

Then, no. of moles,

⇛ No. of moles = 12.7 / 254

⇛ No. of moles = 0.05 moles

⚘ No. of moles of Iodine molecule in the given weight = <u>0.05</u><u> </u><u>moles </u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

6 0

3 years ago

Read 2 more answers