Polyatomic ions: 
, 
, 
, 
, 
, and 
Monatomic ions: 
, 
, and 
<h3>Monoatomic vs Polyatomic Ions</h3>
In chemistry, monoatomic ions are ions that consist of only a single type of atom. They are usually positive or negatively charged and are otherwise known as simple ions. Examples include , , and
Polyatomic ions, on the other hand, are ions that consist of more than one atom, unlike monoatomic ions. The two or more atoms are covalently bonded and the entire structure behaves like a single chemical entity in reactions. Polyatomic ions are otherwise known as molecular ions.
Examples of polyatomic ions are , , , , , and
Thus, from the diagram:
- Polyatomic ions: , , , , , and
More on ions can be found here: brainly.com/question/14982375
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Answer:
36365.4 Joules
Explanation:
The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C), and change in temperature (Φ)
Thus, Q = MCΦ
Since, M = 45.4 g
C = 3.56 J/g°C,
Φ = 250°C - 25°C = 225°C
Q = 45.4g x 3.56J/g°C x 225°C
Q= 36365.4 Joules
Thus, 36365.4 Joules of heat energy is released when the lithium is cooled.
May i please have a(n) answer choices please because it would be a lot better if it was like that and then ill answer it
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
.
