Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
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
Missing question: Express the salt concentration in kg/m³.
Answer is: the salt concentration is 9.8 kg/m³.
m(NaCl) = 9.8 g ÷ 1000 g/kg.
m(NaCl) = 0.0098 kg.
V(solution) = 1 L = 1 dm³.
V(solution) = 1 dm³ ÷ 1000 dm³/m³.
V(solution) = 0.001 m³.
d(solution) = m(NaCl) ÷ V(solution).
d(solution) = 0.0098 kg ÷ 0.001 m³.
d(solution) = 9.8 kg/m³.
