Answer: theres no image or claim
Answer:G
Explanation:
I would look it up on safari just to be sure tho, hope this helps
Answer:
acids
Explanation:
HNO3 is a strong acid (Nitric Acid)
CH3COOH is a weak acid (Acetic Acid
Answer:
SN2
Explanation:
The first step of ether cleavage is the protonation of the ether since ROH is a better leaving group than RO-.
The second step of the reaction may proceed by either SN1 or SN2 mechanism depending on the structure of the ether. Methyl and primary ethers react with HI by SN2 mechanism while tertiary ethers react with HI by SN1 mechanism. Secondary ethers react with HI by a mixture of both mechanisms.
Dipentyl ether is a primary ether hence when treated with HI, the reaction with HI proceeds by SN2 mechanism as explained above.
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>
