We are given the rate law, so we can substitute the given values for the rate constant and the concentrations of the reactants to solve for the rate of reaction. Since rate = k [NH4+][NO2-]:
rate = (2.7 x 10^-4 / M-s)(0.050 M)(0.25 M) = 3.375 x 10^-6 M/s
Answer:
Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
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.
Answer:
1.32 moles.
Explanation:
From the question given above, the following data were obtained:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Number of mole of Al =.?
Next, we shall determine the mass of Al.
This can be obtained as follow:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Mass of Al =?
Density = mass / volume
2.7 = mass of Al / 13.2
Cross multiply
Mass of Al = 2.7 × 13.2
Mass of Al = 35.64 g
Finally, we shall determine the number of mole of Al. This can be obtained as follow:
Mass of Al = 35.64 g
Molar mass of Al = 27 g/mol
Number of mole of Al =?
Mole = mass / molar mass
Number of mole of Al = 35.64 / 27
Number of mole of Al = 1.32 moles
Thus, 1.32 moles of aluminum are present in the block of the metal.
