Answer:
<h2>- It could be stretched into a thin wire.</h2>
Explanation:
As per the question, the most rational claim that the student can make about the aluminum metal is that 'it could be stretched into a thin wire' without breaking which shows its ductility. It is one of the most significant characteristics of a metal. Metals can conduct electricity in any state and not only when melted. Thus, option A is wrong. Options C and D are incorrect as metals neither have the same shape always nor do they break on hitting with a hammer. Therefore, <u>option E</u> is the correct answer.
Answer:
Option (d) is correct
N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺
Explanation:
Total electrons for all the species = 10
So these are <u>iso electronic</u> with each other.
We know
Ionic radii ∝ 
- Si⁴⁺ has 14 protons and 10 electrons
- Mg²⁺ has 12 protons and 10 electrons
- N³⁻ has 7 protons and 10 electrons
- F⁻ has 9 protons and 10 electrons
- Iso electronic species with greatest number of protons have small size and vice versa.
- So Si⁺⁴ have smallest size and N³⁻ have largest in size
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 : The correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.
Explanation :
In the chemical reaction, some substances react together are called reactant and the substance are formed are called product.
During the chemical reaction, the atoms of reactants rearranged to make products. There are on atoms are added or taken away in the reaction. This is known as the conservation of atoms.
For example : carbon atom react with the oxygen to form carbon dioxide.
From the given diagram, we conclude that the arrangement of molecules are different on both side of the mixture of reaction.
On the reactant side, the red molecules bonded with red molecule and the black molecule with white molecules. On the other hand i.e product side, the red molecule bonded with black molecule and white molecule bonded with red molecules. The molecular arrangement are different on both side of the reaction mixture.
Therefore, the correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.
Answer: They always have the same functional groups.
Explanation:
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