Answer:
See explaination
Explanation:
The Cys3-cys97 and cys21-cys142 disulfides restrict the unfolded state of lysozyme enzyme to a class of more compact structures with a less exposed hydrophobic surface, compared to the unfolded states of reduced/non-crosslinked lysozyme. there are 2 major factors which lead to the stabilization of lysozyme due to disulfide bonds-
1- increase in the loop size due to the formation of disulfide bonds that leads to an increase in the even entropic effect.
2- the region formed should be flexible. the strain energy due to the formation of the disulfide bond is lower.
cys21-cys142 has a higher Tm than the cys3-cys97 because it involves flexible parts of the molecule. 21 and 142 residues are located on opposite sides of the active-site cleft where significant hinge-bending motion is seen. this introduces minimal strain in the protein.
1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
Answer:
7mL of sterile water is the initial amount of the concentrated solution is 3mL
Explanation:
In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:
5mg/mL / 1.5mg/mL = 3.33 times
If the initial amount of the drug in the vial is 3mL, the final volume must be:
10mL
That means the volume of water that should be added is:
10mL - 3mL:
<h3>7mL of sterile water is the initial amount of the concentrated solution is 3mL</h3>
The concentration of HCl is equal to 2.54mol/L.
<h3>Mole calculation</h3>
To solve this question, one must use the molarity calculation, which corresponds to the following expression:
Thus, to find the molarity of the sample, the following calculations must be performed:
So, 0.00254 moles were added per 10ml, so we can do:
So, the concentration of HCl is equal to 2.54mol/L.
Learn more about mole calculation in: brainly.com/question/2845237
The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
