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
<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27 27.977 92.23; 28 28.976 4.67; 29 29.974 3.10; abundance =28.07 Silicon
I hope this helps!
Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles =
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
Answer:
Explanation:
we are given
there are in 0.073 gallons
we can see that there are two non-zero places after decimals
so, there are two siginificant figures
step-1: Change gallons to quarts
we know that
1 gallon = 4 quarts
so, we can write as
step-2: Change into liter
we know that
1 liter = 1.057 quarts
or
step-3: Change into millilitre
we know that
Since, it is asking for only two significant figures
so, we can also write as
...................Answer