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:
ACTIVITY 1
Sample 1 has a stronger taste of lemon, and is more sour.
Sample 2 has a sweeter taste, my guess is because there's more sugar:lemon juice ratio.
Answer:
The molar solubility of nitrogen gas is
.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:

where,
= Henry's constant = 
= partial pressure of nitrogen gas
Total air pressure = P = 2.71 atm
Percentage of nitrogen in air = 78.09%
Mole fraction of nitrogen ,

Putting values in above equation, we get:

Hence, the molar solubility of nitrogen gas is
.
Answer:
coffee is a homogeneous mixture while water is a heterogeneous mixture
Answer: A) Ar
Explanation: Argon [Ar] has an atomic no of 18 and the electronic configuration is:
Ar :
Chlorine [Cl] has atomic no of 17 and thus the electronic configuration is :
Cl :
Magnesium [Mg] has atomic no of 12 and thus the electronic configuration is :
Mg :
Sodium has atomic no of 11 and thus the electronic configuration is :
Cl :
As we move across a period the number of electrons are being added to the same shell but the number of protons increases, thus the electrons are being tightly held and thus the atomic radius keeps on decreasing.
Thus the smallest atom would have a tightly bound electron nearer to the nucleus. As argon is the last element of the period, it is the smallest in the period and thus largest amount of energy will be required to remove the outermost electron.