Given:
0.607 mol of the weak acid
0.609 naa
2.00 liters of solution
The solution for finding the ph of a buffer:
[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25
pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.
Answer:
2.01V ( To three significant digits)
Explanation:
First we show the standard reduction potentials of Cu2+(aq)/Cu(s) system and Al3+(aq)/Al(s) system. We can clearly see from the balanced redox reaction equation that aluminium is the anode and was the oxidized specie while copper is the cathode and was the reduced specie. This observation is necessary when substituting values of concentration into the Nernst equation.
The next thing to do is to obtain the standard cell potential as shown in the image attached and subsequently substitute values of concentration and standard cell potential into the Nernst equation as shown. This gives the cell potential under the given conditions.
Radius of Xenon = 1.3Ă—10â’8 cm
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288
Potential energy increases with increase in height as in P.E = mgh so as he runs up, their is more force trying to take it down to a lower ground.