Answer:
57 grams of H3PO4
Explanation:
M= moles/ liters
convert mL to L
234 mL x 1L/1000mL = 0.234L
Rearrange the Molarity formula to solve for moles.
moles= MxL
moles= 2.5M x 0.234L
moles= 0.585 mol
Use the molar mass of H3PO4 to get to grams
0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4
round to two sig figs for 57 grams
Answer:
At physiological pH,the carboxylic acid group of an amino acid will be deprotonated while the amino group will be protonated,yielding the zwitter ion form.
Explanation:
Deprotonated means removal of protons in an acid base reaction and protonated means addition of protons in an acid base reaction.
Both protonated and de protonated reaction takes place in catalytic acid bade reaction by changing either it's mass or it's charge.
During formation of zwitter ion, the carboxylic acid will be deprotonated by donating the H+ ion while the amino acid is protonated by taking the H+ ion.
R-CH-COOH R-CH-COO-
I ⇒ I
NH₂ NH₃ (Zwitter ion)
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is 
I’m pretty sure it’s gravity man