Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:
where, 
is initial molarity and 
is the molarity after dilution. Similarly, 
is the volume before dilution and 
is the volume after dilution.
Let's plug in the values in the equation:
Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.
Answer: it will increase the strong nuclear force in the nucleus
Explanation: Labster
A molecule with two strong bond dipoles can have no molecular dipole if the bond dipoles cancel each other out by pointing in exactly opposite directions. For example, in carbon dioxide (a linear molecule), the carbon-oxygen bonds have a <span>large dipole moment. However, because one dipole points to the left and the other to the right the dipole is cancelled.</span>
Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
