For this problem, we use the Faradays law to relate the mass electroplated and the current passed through. We do as follows:
60.1 g (1 mol / 196.67 g) (1 mol e- / 1 mol Au) ( 96500 C / 1 mol e- ) = 29486.25 C
Q = It
29486.25 = 5t
t = 5897.85 s = 98.30 min
Hope this answers the question.
The tire pressure is the sum of gauge pressure and the standard pressure of atmospheric pressure
We know that the standard air pressure of surrounding = 101.325 kPa
The given gauge pressure = 413 kPascal
So the tire pressure = 101.325 kPa + 413 = 514.325 K Pa
We can also convert the given pascals to atmosphere or Torr or mmHg or bar
That is the volume of the nitrogen atom
2H2 + O2 ---->2H2O
number of moles in reaction 2 mol 1 mol 2 mol
number of liters in the reaction 2*22.4 L 1*22.4 L 2*22.4L
We can see that volumes of the gases are proportional to coefficients in the reaction ( if gases are under the same conditions), so we can write
2H2 + O2 ---->2H2O
2 L 1 L 2 L
given 40 L ( 25 L) 40 L
We can see that we have excess of O2,
because if 2 L H2 are needed 1 L O2, then 40 L of H2 are needed 20 L O2.
So, limiting reactant is H2, and we will need to calculate Volume of H2O using H2.
2L H2 give 2L H2O(gas), so 40 L H2 give 40 L H2O.
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g
Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:
M / E = 4/3
If we cross multiply,
3M = 4E
Now, substituting the value of M from earlier and solving for E,
E = (3 * 0.3357) / 4
E = 0.2518
There are 0.2518 grams of hydrogen per gram of carbon in ethane.