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Leona [35]
2 years ago
6

A 42.0 mL sample of an NaOH solution is placed in a flask with a few drops of phenolphthalein (indicator). If 11.7 mL of a 0.325

M HCl solution is needed to reach the endpoint, what is the molarity of the NaOH solution?
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)
Chemistry
1 answer:
olganol [36]2 years ago
8 0
Please like if you get the answer :)

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professor190 [17]

Answer:

a

Explanation:

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3 years ago
How are catapults an example of a third class lever? Please explain!
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I hope that you can understand this!!! Lol

The thing that you have to pull back to release with, that would be considered a third class lever.

I hope this helps. :)
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3 years ago
Fill up the blank :- If you are using the unit Kelvin, you are measuring the average ......
EastWind [94]
<span>If you are using the unit Kelvin, you are measuring the average </span>kinetic energy per particle. Kelvin is one of the 7 base units in the ISU and is symbolized with a K.

5 0
3 years ago
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The products in a decomposition reaction _____.
alex41 [277]
I believe the answer is  <span>can be elements or compounds
In this case, elements in the decomposition reaction is the substance that cannot be separated into simpler substances.
Compounds, technically act as a reactant in the decomposition reaction, but since the reaction breakdown one substance into two or more, sometimes it exists in the product.</span>
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3 years ago
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In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit
Oliga [24]

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.

3 0
3 years ago
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