Thomson's atomic model is a theory about the atomic structure proposed in 1904 by Thomson, who discovered the electron in 1897, a few years before the discovery of the proton and the neutron.
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Answer:
B: Increasing the volume inside the reaction chamber
Explanation:
PH = -log [H+]
pH = -log (1.0x10^-4) = -(-4) = 4 or A
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
This is a straightforward question related to the surface energy of the droplet.
<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>
