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laila [671]
3 years ago
15

Ruth is 1.23m tall lee is 6 cm shorter than ruth. how tall is lee

Physics
1 answer:
charle [14.2K]3 years ago
7 0
Ok so 1.23m is 123cm, so now all you do is do 123-6=117

i hope this helped

please mark me as brainlest
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If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the c
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<u>Answer:</u> The cell potential is 0.712 V

<u>Explanation:</u>

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

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V

<u>Reduction half reaction:</u> Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V   ( × 2 )

<u>Net reaction:</u> H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)

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-(0.00)=0.80V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +0.80 V

n = number of electrons exchanged = 2

[Ag^{+}]=0.0115M

[H^{+}]=0.355M

Putting values in above equation, we get:

E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V

Hence, the cell potential is 0.712 V

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Explanation:

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