C. Tripling the length and reducing the radius by a factor of 2 is the change to a pipe would increase the conductance by a factor of 12.
<u>Explanation:</u>
As we know that the resistance is directly proportional to the length of the pipe and it is inversely proportional to the cross sectional area of the pipe.
So it is represented as,
R∝ l /A [ area is radius square]
So k is the proportionality constant used.
R = kl/A
Conductance is the inverse of resistance, so it is given as,
C= 1/R.
R₁ = kl₁ / A₁
R₂ = kl₂/A₂
R₂/R₁ = 1/12 [∵ conductance is the inverse of resistance]
= l₂A₁ / l₁A₂
If we chose l₁/l₂= 3 and A₂/A₁= 4 So R₂/R₁= 1/3×4 = 1/12
So tripling the length and reducing the radius by a factor of 2 would increase the conductance by a factor of 12.
At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.
Answer:
I think it both physical & chemical change :')
