THE The answer is B MARK BRAINIEST
Explanation:
Noble gas configuration is defined as the configuration which contains completely filled orbitals.
For example, oxygen atom when gain two electrons then it forms oxygen ion (
).
Atomic number of oxygen atom is 8 and so, its number of electrons will also be 8. But when it gain two electrons then it has total 10 electrons. Hence, electronic configuration of is as follows.
Since, there are completely filled orbitals in an ions. Therefore, it means this ion has a noble-gas configuration.
Thus, we can conclude that any specie which shows completely filled orbitals will have noble-gas configuration.
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:
92.26% of C
Explanation:
To solve this problem we must assume we have 1 mole of benzene. The mole contains 6 moles of C and 6 moles of H. We have to convert these moles to grams in order to find the total mass and mass percent will be:
Mass atom / Total mass * 100
<em>Mass C: </em>6mol C * (12.0107g / mol) = 72.0642g
<em>Mass H: </em>6mol H * (1.00794g / mol) = 6.04764g
<em>total mass: </em>72.0642g + 6.04764g = 78.11184g
Mass percent of C will be:
72.0642g C / 78.11184g* 100
<h3>92.26% of C</h3>
