Answer:
ΔU = 25.8 J
Explanation:
The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.
The work (W) of expansion can be calculated using the following expression:
W = -P. ΔV
where,
P is the external pressure
ΔV is the change in volume
W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J
The change in the internal energy (ΔU) is:
ΔU = Q + W
ΔU = 33.3 J + (-7.48 J) = 25.8 J
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
Chemical reactions can be identified when there is a change in color, energy is produced, change in odor, or if new substance forms.
Answer:
C. 1 cubic foot of loose sand
Explanation:
For many objects having equal volume , surface area will be maximum
of the object which has spherical shape .
But when a sphere is broken into tiny small spheres , total surface area of all the small spheres will be more than surface area of big sphere .
Hence among the given option , surface area of loose sand will have greatest surface area . Loose sand is equivalent to small spheres .
