Perchloric acid (HCO₄) is one of strongest inorganic acids. Perchloric acid precipitation is used to removes most of the protein present in the sample and stabilize many of the small molecule analytes. It can use also to precipitation glycogen, ATP, glutathione, antioxidants.
Perchloric acid donates protons to macromolecules and lower their solubility in water. Also perchloric acid is strong oxidizing agent and change oxidation number of elements.

7 0

3 years ago

Which area would you expect to have more rain? An area that has very low humidity An area that is located near mountains An area

iogann1982 [59]

An area with a very high humidity because the humidity is so much the condenastion going into the clouds cant hold all the evaporation thats happening so fast

8 0

3 years ago

Consider the reaction: 2 H2O (g)-->2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag

denis-greek [22]

DE = dH - PdV

2 H2O(g) → 2 H2(g) + O2(g) 

You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. 

1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. 

Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas 

2 moles reacts to form 3 moles 

The gas equation is 

PV = nRT 
P = pressure 
V = volume (unknown) 
n = moles (1) 
R = gas constant = 8.314 J K^-1 mol^-1 
- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) 
T = temp in Kelvin (kelvin = deg C + 273.15 
So T = 403.15 K 

Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say 

dE = dH -dnRT (because PV = nRT) 

Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules 

dH = 483.6 kJ/mol = 483600 Joules/mol 

dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) 
dE = 483600 J/mol - 3351.77 J 
dE = 480248.23 J/mol 
dE = 480.2 kJ/mol

5 0

3 years ago

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