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Artyom0805 [142]
2 years ago
8

Please help ASAP thank you!!!!!

Chemistry
1 answer:
natali 33 [55]2 years ago
6 0

Answer:

The new volume is 66.3 mL

Explanation:

Boyle's law states that pressure and volume are inversely proportional, and that P1V1=P2V2

P1 = 0.428 atm

P2=1.55 atm

V1 = 240 mL

V2 = ??

(0.428)(240)=(1.55)(V2)

102.72 = 1.55x

x = 102.72/1.55 = 66.27 mL.

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Use the following half-reactions to construct a voltaic cell:
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<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

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What is the daughter nucleus produced when 213Bi undergoes alpha decay?
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