Answer:
3.824 atm
Explanation:
From the ideal gas equation
P = mRT/MW × V
m is mass of testosterone = 12.9 g
R is gas constant = 82.057 cm^3.atm/mol.K
T is temperature of benzene solution = 298 K
MW is molecular weight of testosterone = 288.40 g/mol
V is volume of benzene solution = 286 ml = 286 cm^3
P = 12.9×82.057×298/288.4×286 = 3.824 atm
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
<h2><em>It is True that every bronsted-lowry acid is also a lewis acid </em></h2>
After 3 half-lives, 125 grams of the parent isotope will remain.
