True because evryone reacts a different way
Answer: THE ANSWER IS OZONE WEEEE a layer in the earth's stratosphere at an altitude of about 6.2 miles (10 km) containing a high concentration of ozone, which absorbs most of the ultraviolet radiation reaching the earth from the sun.
Explanation:
The Henderson-Hasselbalch approximation is for conjugate acid-base pairs in a buffered solution. We're going to call HA a weak acid, and A- its conjugate base. The equation is as follows:
pH = pKa + log([base]/[acid]), where the brackets imply concentrations
Plugging in our symbols and the pKa value, the equation becomes:
pH = 4.874 + log([A-]/[HA])
Part 1 : a black dwarf is a theoretical stellar remnant part: 2 idk
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>
