The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is applied to a solution that contains gold in the +3 oxidation state. Calculate the time required to deposit an even layer of gold 1.2 x 10 -3cm thick on the object.

Time required to deposit an even layer of gold with given thickness is sec.

Explanation:

The given data is as follows.

     Surface area = 49.8 ,

     Density of gold = 19.3 ,

     Current = 3.15 A,       thickness of gold layer =

It is known that relation between volume, area and thickness is as follows.

           V = Surface area × Thickness

               =

               = 0.05988

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

        = sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is sec.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.