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
Answer: There are 
atoms of hydrogen are present in 40g of urea, 
.
Explanation:
Given: Mass of urea = 40 g
Number of moles is the mass of substance divided by its molar mass.
First, moles of urea (molar mass = 60 g/mol) are calculated as follows.
According to the mole concept, 1 mole of every substance contains 
atoms.
So, the number of atoms present in 0.67 moles are as follows.
In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.
Thus, we can conclude that there are atoms of hydrogen are present in 40g of urea, .
Answer:
Please see the complete formt of the question below
Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.
MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)
According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.
The answer to the above question is
The limiting reactant is the MnO₂
Explanation:
To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂
Molar mass of MnO₂ = 86.9368 g/mol
Molar mass of HCl = 36.46 g/mol
From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂
However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
1. To step
2. To throw
3. Measure
4. To join
5. Place
