Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
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
Density of liquid= 
.
so, density of liquid= 
= 1.2 gm/cm³.
Answer:
Explanation:
Hello,
In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:
Thus, solving for the moles of the acid, we obtain:
Then, by using the mass of the acid, we compute its molar mass:
Regards.
Answer:
9.86*10^(-3) g
Explanation:
PbSO4 ----> Pb^(2+) + SO4^(2-)
s s
Ksp = s²
s =√Ksp = √(1.8*10^-8) = 1.3*10^(-4) mol/L
The molar solubility PbSO4 = 1.3*10^(-4) mol/L.
2.50 *10^2 mL *1L/10³mL =0.250L
1.3*10^(-4)mol/L *0.250L*303.3 g/mol = 9.86*10^(-3) g
