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

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e