Answer:
22.95g
Explanation:
Given parameters:
Percentage by mass of silver in silver iodide = 45.9%
Mass of silver iodide given = 50g
Unknown:
Mass of silver in silver iodide = ?
Solution:
The formula of the compound is;
Silver iodide = AgI
Since the percentage by mass of silver in the compound is 45.9%;
Mass of silver = percentage by mass of silver x mass of AgI
Mass of silver = 
x 50 = 22.95g
methanol:
1 mole CH3 OH --> produces --> 1 mole CO2
1 mole CO2 has a molar mass of 44.01 gh/mole
your set up is:
(44.01 g CO2) / -726.5kJ = 0.06058g
your answer 0.06058 grams of CO2 produced per kJ released.
Answer: A bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.
Explanation:
A chlorine atom being more electronegative in nature is able to attract a hydrogen atom more readily towards itself as compared to a bromine atom.
Since bromine is less electronegative in nature so bromine will be more selective as a hydrogen abstracting agent. As a result, bromine radical is more stable in nature than chlorine radical.
Thus, we can conclude that bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.
Yes they are only allowing you to break the rule, you would be the exception
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.
