Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
Answer:
attached below
Explanation:
Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum
a) Acetone CH₃COCH₃
Attached below is the structure
b) But-2-yne (CH₃C)₂
Attached below is the structure
To find them you would have numbers of the elements in percentage or grams then you divide them by their molar mass to get their moles. From there you divide by the smallest number. Round it to two or one sig fig. If you have a number that is for ex. 2.5 you multiply it by 2 to make it whole as well the other whole numbers. Then to find the molecular formula the problem must give you another molar mass and using your empirical formula convert it to its molar mass then you divide them, larger number over smaller number. You should get a number round it to 1 sig fig. Now you use that number and multiply the subscripts on the empirical formula to get the molecular formula.
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
