T<span>his is a straightforward question related to the surface energy of the droplet. </span>
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
Right angle acute angle obtuse angle
Answer:
Explanation:
From the 1 g sample you have:
0.546 grams of cobalt chloride
1-0.546=0.454 grams of water
Now:
1) The salt
Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):
Dividing:
So the molecular formula will be:
2) The water
The water's molecular weight is M=18 :
Bonding with the Co:
The complete formula of the hydrate:
Answer:
Freezing T° of solution = - 4.52°C
Explanation:
ΔT = Kf . m . i
That's the formula for colligative property about freezing point depression.
Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.
Li₂O → 2Li⁺ + O⁻²
3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.
m is molality → 0.811 m, this is data
Kf →Cryoscopic constant, for water is 1.86 °C/m
and ΔT = Freezing T° of pure solvent - Freezing T° of solution
We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3
Freezing T° of solution = - 4.52°C
Answer:
activation energy hope this is right
