Answer:
This is called negative adsorption. Hence if the concentration of the adsorbate is less on the surface of the adsorbent than in the bulk, it is known as negative adsorption.
Explanation:
Hope this helps
P=nRTV
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Answer:
yes
Explanation:islands and mountains
Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>
![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>
Answer:
0.8 mL of protein solution, 9.2 mL of water
Explanation:
The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:
C₁V₁ = C₂V₂
We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.
V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL
Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:
(10 mL - 0.8 mL) = 9.2 mL
