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3 years ago

Approximately how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

2 answers:

8 0

<span>c. 30 times The Richter scale is a logarithmic scale for the energy release during an earth quake. For every increase by 1 in magnitude, the energy released is about 31.6 times larger. So calculate the magnitude difference by subtracting. 6.5 - 5.5 = 1.0 And then multiply by 31.6, giving 31.6. And since we're just looking for approximate values, select the answer that's closest which is option "c. 30 times"</span>

Elanso [62]3 years ago

4 0

Answer: (C) 30 times

Richter scale is used to determine the magnitude of an earthquake from the arrival time of P and S waves. It determines the total amount of energy released during an earthquake.
If Richter magnitude scale measures 6.5 then it produces 30 times more energy than 1 Richter scale magnitude during an earthquake. The magnitude of an earthquake is plotted on a logarithmic scale from 1 to 9 and increases 10 folds by one magnitude. If 6.5 is the magnitude, it means it is 10 times more than 5.5 magnitude earthquake. Similarly the amount of energy is calculated as 30 folds. So 6.5 magnitude will have 30 folds more energy released than that with 5.5 magnitude earthquake.

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Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

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From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

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

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

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