The new volume of the dilute solution is 0.33 L.
<u>Explanation:</u>
Using the law of Volumetric analysis, we can find the volume of the dilute solution from the stock solution or the concentrated solution of Calcium Chloride.
V1M1 = V2M2
V1 be the volume of the stock solution = 0.25 L
M1 being the molarity of the stock solution = 0.98 M
V2 be the volume of the dilute solution = ?
M2 being the molarity of the dilute solution = 0.74 M
We have to rearrange the above equation to find V2 as,
V2 = 
Now plugin the values as,
V2 = 
= 0.33 L
So the new volume of the dilute solution is 0.33 L.
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
<span>Draw the Lewis structure. Bromine has 3 bonds and two lone pairs for a trigonal bipyramidal electron geometry and an sp^3d hybridization. Fluorine is peripheral it does not require hybridization, but we often consider it to be hybridized too - it has 1 bond and 3 lone pairs for sp^3 hybridization. So the sigma bonds come from an overlap of an sp^3d orbital on Br with an sp^3 orbital on F. If you don't consider the F to be hybridized the overlap would have to be to a p orbital on the F</span>
Answer:
3.03 g
Explanation:
The first thing to do here is figure out the chemical formula for aluminium hydroxide.
Aluminium is located in group
13
of the periodic table, and forms
3
+
cations,
Al
3
+
. The hydroxide anion,
OH
−
, carries a
1
−
charge, which means that a formula unit of aluminium hydroxide will look like this
[
Al
3
+
]
+
3
[
OH
−
]
→
Al
(
OH
)
3
Now, you can figure out the mass of hydrogen present in
1
mole of aluminium hydroxide by first determining how many moles if hydrogen you get in
1
mole of aluminium hydroxide.
Since
1
mole of aluminium hydroxide contains
3
moles of hydroxide anions, which in turn contain
1
mole of hydrogen each, you can say that you will have
1 mole Al
(
OH
)
3
→
3
a
moles OH
−
→
3
a
moles H
The problem tells you that the molar mass of hydrogen is equal to
1.01 g mol
−
1
. This means that
1
mole of hydrogen has a mass of
1.01 g
.
You can thus say that one mole of aluminium hydroxide contains
3
moles H
⋅
1.01 g
1
mole H
=
a
a
3.03 g H
a
a
∣
∣
I'll leave the answer rounded to three sig figs. Btw my sister calculated this oof
