<span>They have similar or common ancestor. The similarities inherited from the common ancestor as in the case of homologous organs. In evolutionary biology a group of organisms share a common descent if they have a common ancestor. There are many other examples like in the case of hemoglobin to prove the theory of common ancestor</span>
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>
Answer:
Carbohydrates or Carbs have sugar molecules! :3 hope this helped you
Explanation:
Answer:
Explanation:
At STP, 1 mole of hydrogen can occupy a volume of 22.4 liters. At STP, 3 moles of hydrogen can occupy a volume of 2×22. 4=44. 8 liters
We are given the mass spectrum data for this compound which has a molecular ion peak of m+ = 91.043 m/z. When we have an m+ peak that is an odd number, that suggests that there are an odd number of nitrogens, in this case we'll assume 1 nitrogen atom to start. Nitrogen has a mass of 14 so we will substract that from our initial value.
91- 14 (1N) = 77 m/z
We are also told that there are carbon, hydrogens and oxygens present, so we will assume there is at least one oxygen which has a mass of 16 and subtract that value.
77 - 16 (1 O) = 61 m/z
Now we will try to get as close as possible to the remaining mass with carbons that has a mass of 12, and fill the remaining mass with hydrogens that have a mass of 1.
61 / 12 = 5
5 x 12 = 60
61 - 60 (5 C) = 1 m/z and this leaves us with 1 H.
The current formula would be C₅HON, but this structure is impossible since we do not have enough hydrogens to satisfy the carbons. So we can try to use 4 carbons instead and fill the rest with hydrogens.
4 x 12 = 48
61 - 48 (4 C) = 13 m/z and this leaves us with 13 H.
The current formula would be C₄H₁₃ON. The most hydrogens we can have in a compound is 2n+2 where n is the number of carbons. So with 4 carbons the most hydrogens we could have is 10. Therefore, our formula has too many hydrogens and also cannot work. So we cannot make up the remaining mass with carbons and hydrogens, therefore, we should add another oxygen before working with carbons and hydrogens.
61 - 16 (1 O) = 45 m/z
45/ 12 = 3.75
3 x 12 = 36
45 - 36 (3 C) = 9 m/z which gives us 9 hydrogens left.
The current formula is now C₃H₉O₂N. To test if this formula works we can calculate the double bond equivalents (DBE), also known as degrees of unsaturation.
DBE = C - H/2 + N/2 + 1 = 3 - (9/2) + (1/2) + 1 = 0
A value of 0 DBE tells us that there are no double bonds in this molecule but that the formula is a possibility so:
C₃H₉O₂N = 91 m/z
