A10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g co2 and 4.91 g h2o upon complete combustion. what is the empirical formula of the compound?

(23.98 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.544881 mol C 
(0.544881 mol C) × (12.01078 g C/mol) = 6.54445 g C 

(4.91 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.54509 mol H 
(0.54509 mol H) × (1.007947 g H/mol) = 0.54942 g H 

10 g total - 6.54445 g C - 0.54942 g H = 2.90613 g O 
(2.90613 g O) / (15.99943 g O/mol) = 0.18164 mol O 

Divide by the smallest number of moles: 
(0.544881 mol C) / 0.18164 mol = 2.9998 
(0.54509 mol H) / 0.18164 mol = 3.0009 
(0.18164 mol O) / 0.18164 mol = 1.0000 
Round to the nearest whole number to find the empirical formula: 
C3H3O 

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.