Answer: -
12.41 g
Explanation: -
Mass of CO₂ = 42 g
Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol
Number of moles of CO₂ = 
= 0.9545 mol
The balanced chemical equation for this process is
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
From the balanced chemical equation we see
12 mol of CO₂ is produced from 2 mol of C₆H₆
0.9545 mol of CO₂ is produced from 
= 0.159 mol of C₆H₆
Molar mass of C₆H₆ = 12 x 6 + 1 x 6 =78 g /mol
Mass of C₆H₆ =Molar mass x Number of moles
= 78 g / mol x 0.159 mol
= 12.41 g
The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
</span>2.6 x 10^-6 C / 1.6 x 10^-19 (C/electron) = 1.625 x 10^13 electrons
Answer:
(C) im pretty sure is the answer
Explanation:
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
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