Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
Answer:
3. 3.45×10¯¹⁸ J.
4. 1.25×10¹⁵ Hz.
Explanation:
3. Determination of the energy of the photon.
Frequency (v) = 5.2×10¹⁵ Hz
Planck's constant (h) = 6.626×10¯³⁴ Js
Energy (E) =?
The energy of the photon can be obtained by using the following formula:
E = hv
E = 6.626×10¯³⁴ × 5.2×10¹⁵
E = 3.45×10¯¹⁸ J
Thus, the energy of the photon is 3.45×10¯¹⁸ J
4. Determination of the frequency of the radiation.
Wavelength (λ) = 2.4×10¯⁵ cm
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm
2.4×10¯⁵ cm = 2.4×10¯⁷ m
Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m
Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:
Wavelength (λ) = 2.4×10¯⁷ m
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
v = c / λ
v = 3×10⁸ / 2.4×10¯⁷
v = 1.25×10¹⁵ Hz
Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.
Answer: theres no image or claim
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
An eclipse i’m pretty sure right? Lunar maybe?
