a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what is the mepirical formula

C₂H₃O

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

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)

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.