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

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)