Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction. ag+(aq)+e−→ag(s) part a silver can be electroplated at the cathode of an electrolysis cell according to the following half-reaction: ag+(aq)+e−→ag(s) what mass of silver plates onto the cathode when a current of 8.5 a flows through the cell for 68 min ?

Mass of silver deposited at the cathode is 37.1g

Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte

Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.

+e− ==> Ag(s)

Given that

Current (I) = 8.5A

Time (t) = 65 *60 = 3900s

Quantity of electricity passed = 8.5*3900 =33150C

Molar mass of Ag= 108g

96500C will liberate 108g

33150C will liberate Xg

Xg= (108*33150)/96500

=37.1g

Therefore the mass of Ag deposited at the cathode is 37.1g.

The distance in kilometer that Ted will go if he runs at the same pace for 285 minutes is 60km

Step-by-step explanation:

It was given that Ted Ted runs 20 km in 95 minutes, so for us to know how many kilometers he runs at the same pace for 285 minutes by using proportion.

For Ted to run 20 km in 95 minutes.

where given distance= 20km

Given time =95minutes

Speed=20/95 .............eqn(1)

To calculate the distance in kilometers Ted will go if he runs at the same pace for 285 minutes, then

Speed= K/225 ...............eqn(2)

Let us represent distance with 'K'

Given time 225minutes

Equating eqn(1) and eqn(2)

Hence the proportion model that Ted could use in modelling this situation is this;

Cross multiply

K×95=20×225

The distance in kilometer that Ted will go if

he runs at the same pace for 285 minutes is 60km