Answer:
Light moves at 300,000 kilometers per second, divide these and you get 500 seconds, or 8 minutes and 20 seconds this is an average number.
Explanation:
D. Making the reactant particles larger
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
An exponential decay law has the general form: A = Ao * e ^ (-kt) =>
A/Ao = e^(-kt)
Half-life time => A/Ao = 1/2, and t = 4.5 min
=> 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154
Now replace the value of k, Ao = 28g and t = 7 min to find how many grams of Thalium-207 will remain:
A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g
Answer 9.5 g.
Answer:Complex carbohydrates are formed from monosaccharides, nucleic acids are formed from mononucleotides, and proteins are formed from amino acids. There is great diversity in the manner by which monomers can combine to form polymers. For example, glucose monomers are the constituents of starch, glycogen, and cellulos
Explanation:
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