<span>It contains 6.02mc001-2 particles of a given substance</span>
The answer is mass which is also m/v=d
An acid is a compound which will give H+ ions or H3O^+ ions
the reaction will be
![[Mn(H_{2}O )_{6} ^{+3} +H_{2}O --> [MnOH(H_{2}O)_{5}]^{+2} + H_{3}O^{+}](https://tex.z-dn.net/?f=%5BMn%28H_%7B2%7DO%20%29_%7B6%7D%20%5E%7B%2B3%7D%20%2BH_%7B2%7DO%20--%3E%20%5BMnOH%28H_%7B2%7DO%29_%7B5%7D%5D%5E%7B%2B2%7D%20%2B%20H_%7B3%7DO%5E%7B%2B%7D)
Thus as there is evolution of H_{3}O^{+} the Mn+3 is an acid
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.