Answer:
fluoride ion with a charge of -1
Explanation:
If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
___________________________________________________
Carbon is used to reduce the oxide of aluminium to get the metal, in case condensation happens during this process.
___________________________________________________
Hope this helps!
Answer: The answer is salt
Answer:
See explanation
Explanation:
the Cis structure, we can have two possibilities. The methyl groups can go both in <u>axial positions</u> or both in <u>equatorial positions</u>. We have to remember that cis it means <u>"same orientation"</u>, so in the axial positions, both methyl groups go up. (Both have the same orientation). In the equatorial positions, both groups go down.
In the axial positions, we will have <u>more steric hindrance</u> because the groups are close to each other. Therefore, we will have <u>more energy</u> and the structure will be <u>less stable</u>. In the equatorial positions, we <u>dont have any steric hindrances</u>, so we will have <u>less energy and more stability</u>.
See figure 1
I hope it helps!