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.
Question 9. The first one is the smallest. Anything with a negative exponent is going to be less than 1, the .00000241. The exponent tells you the number of zeroes to the right of the decimal point. Farther to right gets smaller and smaller.
Question 10. The last one is true. If the last digit is smaller than 5, drop the digit, and do not change. (If it is a 5 or larger, the digit before it would round up)
NO alkenes can not react with potassium dichromate
On point? Do you have any options?