Answer:
2.24 Liters are in 4.4 grams of CO2 at STP
5) state
when the state of the chemical changes that is the most important
The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.
The heat energy is given as :
q = m c ΔT + Ccal ΔT
q = ( 974 g× 4.184 ×6.9) + 624 ×6.9
q = 32424.59 J
moles of hydrocarbon = 0.0418 mol
heat of combustion = 32424.59 J / 0.0418 mol
= 775707.89 J
= 775.70 kJ
Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.
To learn more about calorimeter here
brainly.com/question/28943378
#SPJ4
Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.
I would say A or C because the other 2 are decided before a baby leaves the womb
