Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer:
8 to 1.
Explanation:
- Oxygen combines with hydrogen atoms to form water according to the balanced equation:
<em>O₂ + 2H₂ → 2H₂O.</em>
It is clear that one mole of oxygen combines with two moles of hydrogen atoms to form 2 moles of water.
So, the molar ratio of oxygen to hydrogen is (1 to 2).
- The mass of 1 mole of oxygen = (no. of moles)(molar mass) = (1 mol)(32.0 g/mol) = 32.0 g.
- The mass of 2 moles of hydrogen = (no. of moles)(molar mass) = (2 mol)(2.0 g/mol) = 4.0 g.
<em>So, the mass ratio of oxygen to hydrogen (32.0 g/4.0 g) = (8: 1).</em>
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Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
