Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
Answer:
* 
* The solution is acidic since the pH is below 7.
Explanation:
Hello,
In this case, we can mathematically define the pH by:
![pH=-log([H_3O^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH_3O%5E%2B%5D%29)
Thus, for the given hydronium concentration we simply compute the pH:

Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.
Regards.
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.