Peptide bonds, hydrogen bonds, disulfide linkages, van der Waals, and electrostatic forces of attraction are the chemical bonds that stabilize the different structures in primary, secondary, tertiary, and quaternary structures.
The advanced structure of proteins gives rise to two kinds of major molecular shapes which are fibrous as well as globular structures. The main forces which are involved in stabilizing the secondary along with the tertiary structures of proteins include hydrogen bonds, disulfide type linkages, van der Waals attraction, and electrostatic forces of attraction.
The primary structure is generally determined by adjoining peptide bonds where the link is adjoining amino acids in sequential order. Tertiary structure is determined by the existence of disulfide bonds in between hydrophobic interactions as well as cysteine residues whereas the quaternary type structure is determined by multiple subunits of a protein that undergo various interactions.
Hydrogen bonds exist in a protein molecule as its large number can form between adjacent regions of the polypeptide chain in folded form and stabilize its three-dimensional kind of shape.
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Answer:
the pressure of gas is 100.0 the volume is 500.0
Explanation:
2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃
m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol
1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw
2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}
m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}
m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}
m(Fe)=55.847*245*0.805/179.857= 61.24 g
<u>Answer:</u> The equilibrium constant for this reaction is 
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-587.4%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%284%5Ctimes%20%28-137.3%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-38.2kJ%2Fmol)
To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:
where,
= Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![58^oC=[273+58]K=331K](https://tex.z-dn.net/?f=58%5EoC%3D%5B273%2B58%5DK%3D331K)
= equilibrium constant at 58°C = ?
Putting values in above equation, we get:
Hence, the equilibrium constant for this reaction is
Answer:
ex.1
1.The water in the test tube expands as it is heated. There is only one direction to expand in, the glass tube. This glass tube is narrow so that a small change in volume will result in a considerable change in height; your signal is amplified. This is the same phenomenon as in thermometers with a glass bulb on the lower end.
2. there will be a thermal expansion will happened.
