Answer:
It's 4.1 moles Li2O = 94 g on Odyssey ware
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:
The SI base unit for amount of substance is the mole. 1 grams NaOH is equal to 0.025001806380511 mole.
Answer:
Option D is correct = 8.12 grams of NaCl
Explanation:
Given data:
Moles of sodium chloride = 0.14 mol
Mass of sodium chloride = ?
Solution:
Formula:
Number of moles = mass of NaCl / Molar mass of NaCl
Molar mass of NaCl = 58 g/mol
Now we will put the values in formula.
0.14 mol = Mass of NaCl / 58 g/mol
Mass of NaCl = 0.14 mol × 58 g/mol
Mass of NaCl = 8.12 g of NaCl
Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.
