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:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
Hi the answer is 1/6, because 1/6 of 3,600 is 600, so it's 1/6. Have a great day!
Answer:
Star's
Explanation:
Stars have their own brightness
0.000001ppm
Explanation:
Mass of fluoride = 500g
Volume of water = 500000liters
Unknown:
Parts per million of fluoride = ?
Solution:
The parts per million is the amount of solute in milligram dissolved in a liter of water or milligram per kilogram of solvent
It is a unit used to express very small concentration.
we need to convert g - mg
500g = 500 x 10⁻³mg = 0.5mg
Concentration in parts per million = 
Concentration in parts per million =
= 0.000001ppm
learn more:
Parts per million brainly.com/question/2854033
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