Valance electron are called
It is given that the surface area of sphere is 4 π r² and its volume is (4/3 π r³)
With a diameter of 1.2 mm you have a radius of 0.6 mm so the surface area about 4.5 mm² and the volume is about 0.9 mm³
The total surface energy of the original droplet is (4.5 x 10⁻⁶ m x 72) = 3.24 x 10⁻⁴mJ
The five smaller droplets need to have the same volume as the original so:
5 V = 0.9 mm³ so the volume of smaller sphere will equal 0.18 mm³
Since this smaller volume still have volume (4/3 π r³) so r = 0.35 mm
Each of the smaller droplets has a surface are = 1.54 mm²
The surface energy of the 5 smaller droplet is then (5 x 1.54 x 10⁻⁶ m x 72) = 5.54 x 10⁻⁴ mJ
From this radius the surface energy of all smaller droplets is 5.54 x 10⁻⁴ and the difference in energy is (5.54 x 10⁻⁴) - (3.24 x 10⁻⁴) = 2.3 x 10⁻⁴ mJ
Therefore we need about 2.3 x 10⁻⁴ mJ of energy to change a spherical droplet of water of diameter 1.2 mm into 5 identical smaller droplets
Answer: 1.15L
Explanation:Please see attachment for explanation
Answer:
18. E. 0%
19. D. 25%
Explanation:
Question 18:
Let's use "B" to represent the dominant allele of "light blue skin", and
"b" for recessive "light green skin".
Squidward => BB - light blue skin
Squidward's bride => bb - light green skin
When they cross, they will have the following offsprings:
(BB) × (bb) - Parent
(Bb) (Bb) (Bb) (Bb) - Offspring
All the offspring would be light green skin. The dominant allele of light green skin will express itself over the recessive allele.
Therefore, the chances of Squidward and his bride having light green skin is 0%
Question 19:
Squidward's son => Bb - light blue skin
Squidward's son's bride => Bb - light blue skin
(Bb) × (Bb) - parent
(BB) (Bb) (Bb) (bb) - offspring
They will have the following offspring:
BB and Bb - light blue - 75%
bb - light green - 25%
Chance of having offspring with light green skin is 25%
Answer : Option C) An induced dipole will be produced in the molecule on the right.
Explanation : As per the given information there is a polar molecule which is placed on the left side which has partial positive charge at one end and on other end has partial negative charge which shows that it has a dipole in it. It tries to induce the non-polar molecule which is at right side. So, there will be an induce dipole interaction between both when they are placed closer to each other.
