Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C
I hope this helps
Answer:
The correct answer is "obligatory water reabsorption in the proximal convoluted tubule".
Explanation:
The mechanism for producing concentrated urine cannot include the obligatory reabsorption of water in the proximal convoluted tubule since this process is part of the nephron, the system that filters the blood. Glucose and amino acids are reabsorbed almost entirely, as are approximately 70% of filtered potassium and 80% of bicarbonate.
Have a nice day!
Answer:
Activation energy of phenylalanine-proline peptide is 66 kJ/mol.
Explanation:
According to Arrhenius equation- 
, where k is rate constant, A is pre-exponential factor, 
is activation energy, R is gas constant and T is temperature in kelvin scale.
As A is identical for both peptide therefore-
![\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bala-pro%7D%7D%7Bk_%7Bphe-pro%7D%7D%3De%5E%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-E_%7Ba%7D%5E%7Bala-pro%7D%5D%7D%7BRT%7D)
Here 
, T = 298 K , R = 8.314 J/(mol.K) and 
So, ![\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.05%7D%7B0.005%7D%3De%5E%7B%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-%2860000J%2Fmol%29%5D%7D%7B8.314J.mol%5E%7B-1%7D.K%5E%7B-1%7D%5Ctimes%20298K%7D%7D)
(rounded off to two significant digit)
So, activation energy of phenylalanine-proline peptide is 66 kJ/mol
Control rods are used<span> in </span>nuclear<span> reactors to </span>control<span> the fission rate of uranium and plutonium. They are composed of chemical elements such as boron, silver, indium and cadmium that are capable of absorbing many neutrons without themselves fissioning.</span>
Pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles
