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alexira [117]
3 years ago
11

Any help is greatly appreciated!!

Chemistry
1 answer:
Marina CMI [18]3 years ago
3 0

Answer:

B) 473 mL

Explanation:

So there was this guy named Charles who discovered the relation between volume and temperature and his law called Charles Law says volume and temperature are directly related that if one goes up the other goes up as well.

It can be summarized using V1/T1= V2/T2 (where 1 is initial condition and 2 is after condition) (V is volume and T is temperature)

Based on your problem, V1 is 568mL, T1 is 298 K and T2 is 248 K ans we are trying to find V2 so its simple because we use the formula I cited earlier.

568ml/298K= V2/248K (divide 568 by 298 and multiply this calculation by 248)

V2= 473 mL

The answer should make sense because in initial condition you were at higher temperature thus your volume was larger but since your after condition is when temperature is lowered to 248K the volume also subsequently lowers going from 568mL to 473 mL

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What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.
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Explanation :

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Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o{cell} of the reaction, we use the equation:

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

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Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}

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n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}

E_{cell}=1.14V

Therefore, the voltage of the voltaic cell is 1.14 V

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3 years ago
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Answer:

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