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Liono4ka [1.6K]
3 years ago
14

What are 2 comparisons between organisms and music?

Chemistry
2 answers:
Brrunno [24]3 years ago
6 0

Answer:

Explanation:

Can I please have this for some reason all my answers got deleted (I had 28)

Now are gone so you don't have to give thanks just let me get out of the negative. (pls!!)

muminat3 years ago
6 0
They can form of inteqaty
You might be interested in
How many moles of silver are there in 6.9 x 10^28 silver atoms?
Juli2301 [7.4K]

Answer:

1.2 x 10⁵ moles Ag (2 sig. figs.)

Explanation:

1 mole any substance (elements or compounds) => 6.023 x 10²³ particles of specified substance

∴ 6.9 x 10²⁸ atoms Ag = 6.9 x 10²⁸ Ag atoms / 6.023 x 10²³ Ag atoms/mole Ag

                                    = 1.145608501 x 10⁵ moles Ag (calculator answer)

                                    = 1.2 x 10⁵ moles Ag (2 sig. figs.)

6 0
2 years ago
Read 2 more answers
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
Now let us refine our model by noting that there is a second source of Na+ ions in the cell: NaI. Suppose the outside of the cel
MArishka [77]

Answer:

E=55mV

Explanation:

Hello,

Considering the given information, the new concentration of Na+inside will be (13.6 + 4) mM = 17.6 mM, and outside will be (150 + 0.04) mM = 150.04 mM due to the previous specified conditions.

Now, the recalculation of the Nerst potential at the supposed temperature of 25 °C (which is modifiable) is done via:

E=\frac{RT}{zF} ln(\frac{C_{Na^+,outside}}{C_{Na^+,outside}} )\\\\E=\frac{8.314\frac{J}{mol*K}*298K}{1*9.65x10^4\frac{C}{mol} } *ln(\frac{150.04}{17.6} )=0.055V*\frac{1x10^3mV}{1V} \\E=55mV

Best regards.

8 0
3 years ago
Which of these did your answer include?
sveticcg [70]

The oxygen atom has a partial negative charge and the Hydrogen atoms have a partial positive charge

8 0
3 years ago
Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv
poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0
3 years ago
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