The answears are in the attached photo.
Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2
Answer:
The complete aerobic oxidation of glucose is coupled to the synthesis of as many as 36 molecules of ATP
Explanation:
Glycolysis, the initial stage of glucose metabolism, takes place in the cytosol and does not involve molecular O2. It produces a small amount of ATP and the three-carbon compound pyruvate. In aerobic cells, pyruvate formed in glycolysis is transported into the mitochondria, where it is oxidized by O2 to CO2. Via chemiosmotic coupling, the oxidation of pyruvate in the mitochondria generates the bulk of the ATP produced during the conversion of glucose to CO2. The biochemical pathways that oxidize glucose and fatty acids to CO2 and H2O.
Its Homogenous Centrifuges are used to speed up the process of separating Homogeneous mixtures.
Answer:
2.12×10²³ atoms.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.
Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:
1 mole of zirconium also 6.02×10²³ atoms.
Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.
Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.
