Answer:
The solution will turn red.
Explanation:
HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺
(red) (yellow)
Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,
At about pH 3.4, the two the forms are present in equal amounts, and the indicator colour is orange.
If you add more acid, you are disturbing the equilibrium.
According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.
The system will try to get rid of the added acid, so the position of equilibrium will move to the left.
More of the unionized molecules will form, so the solution will turn red.
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:
C) H2S
Explanation:
In chemistry, the dissolution of one substance in another is dependent on the magnitude of intermolecular interaction between the two substances. Hence, if two substances do not interact in one way or the other, then one can not dissolve the other.
Let us consider the fact that NH3 is a polar molecule and it is a general principle that like dissolves like. Hence, only H2S which is also a polar molecule can effectively interact with NH3 due to dipole-dipole interaction between the two molecules.
Also, ammonia reacts with hydrogen sulphide as follows;
2NH3 + H2S → (NH4)2S
Hence H2S is more likely to dissolve in NH3.
Answer:
Ba(ClO₃)₂ → BaCl₂ + 3 O₂
Explanation:
When exposed to heat, barium chlorate (Ba(ClO₃)₂ breaks down into an inorganic compound (Barium chloride - BaCl₂) and a molecule (Oxygen - O₂).
