Here we have to draw the four isomers of the compound 3-bromo-4-fluorohexane.
The four isomers of the compound is shown in the figure.
In an organic molecule the chiral -C center is that where four (4) different groups are present. In 3-bromo-4-fluorohexane the 3 and 4 positions are chiral centers. The possible isomers of a molecule can be obtained from the formula 2n. As here 2 chiral centers are present thus number of stereoisomers will be 2×2 = 4.
The four different isomers as shown in the figure are 3R-, 4R-; 3S-, 4S; 3R, 4S and 3S-, 4R- 3-bromo-4-fluorohexane.
In the 3-bromo-4-fluorohexane the functional groups are -Br, C₂H₅, -C₃H₆F and -H for 3-position and -F, -C₂H₅, -C₃H₆ and -H for 4-position respectively.
The priority of the -3 position will be Br > C₃H₆F > C₂H₅ > H and for -4 position F > C₃H₆Br > C₂H₅ > H. If the rotation from the higher priority group to lower is clockwise and anticlockwise then the S- and R- notation are used respectively. However if the -H atom is present at the horizontal position then the notation will be reverse.
Thus the four isomers of the compound is shown.
Just use the Heisenberg Uncertainty principle:
<span>ΔpΔx = h/2*pi </span>
<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>
<span>Hint: </span>
<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>
<span>Δp = mv*(0.01) </span>
<span>m = mass of electron </span>
<span>v = velocity of electron </span>
<span>Solve for Δx </span>
<span>Δx = h/(2*pi*Δp) </span>
<span>And that is the uncertainty in position. </span>
The first option, collapsed in on itself.
The star's core mass becomes so dense that the resulting gravity implodes the star.
Interesting enough, the third option is kindof true too...some large and tenacious black holes that absorb other stars will form incredibly bright accretion disks around their perimeter before filling absorbing the star.
Answer:
Las bebidas gaseosas como las gaseosas están hechas de un soluto de dióxido de carbono gaseoso en un líquido. La solubilidad del dióxido de carbono en el líquido depende de la presión y la temperatura de la lata de refresco, y también de agitar la lata de refresco que introduce burbujas que permanecen ocultas hasta que se abre la lata antes de que burbujee.
Por lo tanto, dado que la presión en la lata de refresco permanece constante, elevar la temperatura, agitar la lata de refresco o congelar el refresco, lo que aumenta la cantidad de dióxido de carbono en la porción líquida, hará que el refresco forme espuma y se derrame.
Explanation:
