How many amperes of current are required to deposit 5.00 g of gold on the cathode in 60.0 minutes from a solution containing a salt of gold (III)?

=      2.04 amps

Explanation:

5.00 g Au = 5.00 g Au X (1.00 mol Au/197 g Au) X (3 mole e⁻/mole Au) X (96,500amp·sec/mole e⁻) X (1min/60sec)  X (1/60min)                                

=      (5.00)(1.00/197)(3.00/1)(96,500/1)(1/60)(1/60)amps

=      2.04 amps

d

explanation:

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol