Answer:
I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong
Explanation:
Answer:
The volume is increased.
Explanation:
According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;
V₁ / T₁ = V₂ / T₂ -----(1)
In statement the data given is,
T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C
T₂ = 20 °C = 293.15 K
So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,
V₁ = 100 L
V₂ = Unknown
Now, we will arrange equation 1 for V₂ as,
V₂ = V₁ × T₂ / T₁
Putting values,
V₂ = 100 L × 293.15 K / 283.15 K
V₂ = 103.52 L
Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.
Answer:
A) Separating funnel method
B) Simple Distillation
C) Evaporation
D) Sublimation
E) It is based on the principle of separation whereby even though two substances are dissolved in the same solvent, their respective solubilities could be different. Thus, the component that has more solubility will rise fastest and will therefore get separated from the mixture.
Explanation:
A)
B) Kerosene and petrol are both miscible liquids and the difference in their boiling point temperature is not more than 25°C. Thus, we make use of Simple distillation.
C) Can be separated by evaporation where the water is boiled and it evaporates and leaves the salt behind
D) To separate camphor from salt, we use sublimation so the camphor can change directly from solid to the gas state without passing through the liquid state.
E) Chromatography is used to separate components of a mixture.
It is based on the principle of separation whereby even though two substances are dissolved in the same solvent, their respective solubilities could be different. Thus, the component that has more solubility will rise fastest and will therefore get separated from the mixture.
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
