Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.
You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume
C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL
C1 x 250 mL = 0.2 M x 35 mL
C1 = (0.2 M x 35 mL) / 250 mL
C1 = 0.028 M of water added to 35mL of 0.2M HCl
Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
Vinegar is the only thing listed
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>
