The given question is incomplete. The complete question is :
Carbon tetrachloride can be produced by the following reaction:
Suppose 1.20 mol 
of and 3.60 mol of 
were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of 
. Calculate equilibrium constant at the unknown temperature.
Answer: The equilibrium constant at unknown temperature is 0.36
Explanation:
Moles of 
= 1.20 mole
Moles of 
= 3.60 mole
Volume of solution = 1.00 L
Initial concentration of = 
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 1.20 M 3.60 M 0 0
At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BS_2Cl_2%5D%5Ctimes%20%5BCCl_4%5D%7D%7B%5BCl_2%5D%5E3%5BCS_2%5D%7D)
Now put all the given values in this expression, we get :
Given :Equilibrium concentration of , x = 
Thus equilibrium constant at unknown temperature is 0.36
Its air and water. if its multiple choice
Answer:
C. More NO2 and SO2 will form
Explanation:
Le Chatelier's Principle : It predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc
It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.
In given Reaction SO3 is introduced(increased) .
So equilibrium will shift in the direction where SO3 should be consumed(decreased)
Hence the equilibrium will go in backward direction , i.e
So more and more Of NO2 and SO2 will form
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.
Accuracy is when you hit as close as to the target as you can and precision is when you are on point
