Answer:
There is no formation of shmoos
Explanation:
If the hypothesis that says "Fus3 kinase is required for the signal transduction pathway leading shmoo formation" is correct, then a strain with a deletion Fus3, is not going to render Fus3 kinase, at least active. By this reason, there will not be shmoo formation
Volume of a sphere = 4/3 x pi x r^3
When put a fraction of volume constant 4/3 x pi cancels out.
So only cube of radii remains.
Radius of proton = 10^-15 m (Fact; remember it)
Radius of total Hydrogen atom = 0.529 × 10^−10 m
Fraction of Volumes : R'^3/R^3 = (R/R)^3
Fraction = ((10^-15)/(0.529 × 10^−10m.))^3= (1/52900)^3 =
6.755 x 10^-15
Is it 8.06?
Or 58.57?
Don't get mad if there wrong!!
But please let me know if it's right or wrong tho.
Answer:
Of the following equilibria, only one will shift to the right in response to a decrease in volume.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Decrease the volume
If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.
The amount of parent and daughter material trapped in the rock.