Answer:
We identify nucleic acid strand orientation on the basis of important chemical functional groups. These are the <u>phosphate</u> group attached to the 5' carbon atom of the sugar portion of a nucleotide and the <u>hydroxyl</u> group attached to the <u>3'</u> carbon atom
Explanation:
Nucleic acids are polymers formed by a phosphate group, a sugar (ribose in RNA and deoxyribose in DNA) and a nitrogenous base. In the chain, the phosphate groups are linked to the 5'-carbon and 3'-carbon of the ribose (or deoxyribose) and the nitrogenous base is linked to the 2-carbon. Based on this structure, the nucleic acid chain orientation is identified as the 5'-end (the free phosphate group linked to 5'-carbon of the sugar) and the 3'-end (the free hydroxyl group in the sugar in 3' position).
Answer:
1250N
Explanation:
This question is based on pascal's Law.
So By Pascal's Law
=
therefore =force on input piston =25N
= Force or weight on output person.
therefore after putting the values we get,
= (25x 1500)/30
=1250N
If one were to match the ratio of atoms of the elements found in this molecular formula of artificial sweetener it would be :
Carbon - 7 atoms
Hydrogen - 5 atoms
Nitrogen - 1 atom
Oxygen - 3 atoms.
Answer:
Risk management can be described as the identification and evaluation of certain risks beforehand and the appropriate steps that can be taken to avoid the problematic situations. Risk management is the forecasting of certain risks and the plans made to resist or overcome these risk situations.
An example of a risk management strategy can be a pet owner understanding that he'll have to make major payments for the health of the pet or will have to pay vet bills. In order to save money, the pet owner decided to go for pet insurance beforehand so that any problematic situation can be avoided.
Answer:
D. To ensure the cooling process is not affected by surrounding temperature
Explanation:
The conical flask acts as a <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>j</u><u>a</u><u>c</u><u>k</u><u>e</u><u>t</u><u>.</u>