Answer:
4 Co(s) + 3 O2(g) = 2 Co2O3(s)
Explanation:
Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
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:
4.94g of material
Explanation:
Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:
Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>
100mL of a 5% solution contains 5g of material in 100mL of water. Thus:
8 = X / 100mL / (5g-X) / 100mL
<em>Where X is the amount of material in grams that comes to the organic phase.</em>
8 = X / 100mL / (5g-X) / 100mL
8 = 100X / (500-100X)
4000 - 800X = 100X
4000 = 900X
4.44g = X
<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>
And will remain 5g-4.44g = 0.56g.
In the second extraction:
8 = X / 100mL / (0.56g-X) / 100mL
8 = 100X / (56-100X)
448 - 800X = 100X
448 = 900X
0.50g = X
<em>In the second extraction, you will extract 0.50g of material</em>
Thus, after the two extraction you will lost:
4.44g + 0.50g = <em>4.94g of material</em>
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The reason why the reaction written on the picture can be classified as a synthesis reaction is :
the reaction shows one compound that formed from two compounds
hope this helps
