<span>Melting is an endothermic process (i.e. it absorbs heat), whereas freezing is an exothermic process (i.e. it releases heat, or can be thought of, albeit incorrectly from a thermodynamics standpoint, as "absorbing cold"). The standard enthalpy of fusion of water can be used for both scenarios, but standard enthalpy is in units of energy/mass, so 10 times as much energy will be absorbed in the former scenario (melting 10 kg of ice) than what will be absorbed in the latter scenario (freezing 1 kg of water). For both processes, assuming the water is pure and at standard atmospheric pressure, and the entire mass remains at thermal equilibrium, the temperature of both the solid and the liquid will remain at precisely 0 degrees Celsius (273 K) for the duration of the phase change.</span>
Answer:
5' RG GWCCY 3'
3' YCCWG GR 5'
Explanation:
The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).
The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'
target Sequence: 5' RGGWCCY 3'
3' YCCWGGR 5'
The enzyme cleavage point is at:
5' RG^GWCCY 3'
3' YCCWG^GR 5'
The product of the cleavage will give a sticky end Cleavage:
5' RG GWCCY 3'
3' YCCWG GR 5'
Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.
Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
Answer : The half-life at this temperature is, 3.28 s
Explanation :
To calculate the half-life for second order the expression will be:
![t_{1/2}=\frac{1}{k\times [A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5Ctimes%20%5BA_o%5D%7D)
When,
= half-life = ?
= initial concentration = 0.45 M
k = rate constant = 
Now put all the given values in the above formula, we get:


Therefore, the half-life at this temperature is, 3.28 s