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
It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341
I’m not too sure I hope someone answers for you
Answer:
N2
Explanation:
Rate of effusion is defined by Graham's Law:
(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))
(Where M is the molar mass of each substance. )
Molar Mass of oxygen, O2, is 32 (M1).
Rate of effusion of O2 to an unknown gas is .935(Rate 1).
Rate 2 is unknown so put 1.
Solve for x (M2).
.935/1 = sqrt x/ sqrt32
.935 x sqrt 32 = sqrt x
5.29 = sq rt x
5.29^2 = 27.975 = 28
N2 has a molar mass of 28 so it is the correct gas.
Answer:
The answer is 1 and 4.
Explanation:
Mass is most concentrated in the nucelus of an atom. Therefore, if you are looking to find the area with the least mass, go outside of the nucelus. Points one and four are the furthest outside of the nucleus.
