<span>Only if they have the same mass.</span>
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:
<u>132.15</u>
Explanation:
Molar mass N = 14.00
Molar mass H = 1.01
Molar mass H4 = 1.01 x 4 = 4.04
Molar mass NH4 = 14.00 + 4.04 = 18.04
Molar mass (NH4)2 = 18.04 x 2 = 36.08
Molar mass S = 32.07
Molar mass O = 16.00
Molar mass O4 = 16.00 x 4 = 64.00
Molar mass SO4 = 32.07 + 64.00 = 96.07
Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>
Ml=−2,−1,0,+1,+2.
<span>Since each of these orbitals can hold a maximum of </span>two electrons<span>, one having spin-up and one having spin-down, a total of </span>10 electrons<span> can share the quantum numbers n = 4 and l = 2</span>
