Answer:
living organisms whose genetic material has been artificially manipulated in a laboratory through genetic engineering
Explanation:
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.
Answer:
phosphodiester bond
Explanation:
<em>Phosphodiester linkage/bond is found in deoxyribonucleic and ribonucleic acids. It is formed from a reaction involving the elimination of water from a reaction involving the hydroxyl groups of two different 5-carbon (pentose) sugars and a phosphate group.</em>
The elimination of water, also known as condensation reaction occur twice, resulting in the formation of two ester bonds which then bind the phosphate group to the pentose sugars to become a phosphodiester bond.
The bond links the 3'-hydroxyl group of one of the pentose sugars and the 5'-hydroxyl group of the other pentose sugar in the nucleotides that make up nucleic acids.
