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:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Answer:
You are the Cobalt
Explanation:
The least massive metalloid in the fourth period is Germanium, and it have 32 protons. If you have 5 less protons: 32 - 5 = 27 protons. The element with 27 protons is Cobalt
Answer:
the radius is half the diameter
The answer is [Ne] 3s^2 3p^5 because chlorine is the fifth element in the 3rd row of elements in in p orbital
