Answer:
To understand the utility in sequence comparison and in the search for proteins that have a common evolutionary origin, you need to be clear about some concepts about how to evolve proteins. The idea that is accepted is that throughout the evolution some species are giving rise to new ones. Behind this is the genetic variation of organisms, that is, the evolution of genomes and their genes, as well as the proteins encoded by them.
Explanation:
Three ways can be distinguished by which genes evolve, and by proteins: mutation, duplication and shuffling of domains. When differences between homologous protein sequences are observed, these differences change to do with the way of life of the organism, an example of this, bacteria that live in hot springs at very high temperatures have proteins with a very high denaturation temperature, and these proteins are usually richer in cysteines. On the other hand, the fact that in positions of the sequences they remain unchanged (preserved positions), means that these have a special importance for the maintenance of the structure or function of the protein and its modification has not been tolerated throughout of evolution
An inert gas will not react with either the reactants or the products, so it will have no effect on the product/reactant ratio, and therefore, it will have no effect on equilibrium.
Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL
Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g
The weight/weight % or percent by mass of the solute is 5.41 %.
Answer:
those have symbols for their Latin or Greek name
Explanation:
hope it helps
<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
