How many moles of O_{2} molecules are present in a 0.301 L sample of dry air at 0.0395 atm and 23.8 °C? (Oxygen constitutes 20.94% of the molecules in dry air.) (Enter your answer using either standard or scientific notation. For scientific notation, 6.02 x 10^{23} is written as 6.02E23.)

The answer to your question is n = 4.89E-4

Explanation:

Data

Volume = 0.301 L

Pressure = 0.0395 atm

Temperature = 23.8 °C

moles = ?

R = 0.082 atm L /mol°K

Process

1.- Convert the temperature to °K

Temperature = 273 + 23.8

                      = 296.8°K

2.- Use the ideal gas law to find the answer

                  PV = nRT

-Solve for n

                    n = PV / RT

-Substitution

                    n = (0.0395)(0.301) / (0.082)(296.8)

-Simplification

                    n = 0.011889 / 24.3376

-Result

                    n = 0.000488 or n = 4.89 x 10⁻⁴ or n = 4.89E-4

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.