Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
The first element to have 4d electrons in its electron configuration is Yttrium with an atomic number of 39 and the chemical symbol Y. The electron configuration of an atom refers to the distribution of its electrons in the atomic orbitals. These orbitals are represented by letters s,p,d,f and so on. The electron configuration of Yttrium is [Kr] 4d¹ 5s², wherein [Kr] refers to the electron configuration of the noble gas, Krypton.
PbO2
You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula would be PbO2
