The arrangement of particles that make up an ionic compound would be an ionic lattice type of crystal arrangement. An ionic lattice type of structure will be formed due to many of the ionic bonds formed between the oppositely charged ions of the metal and nonmetal.
A mole of an element refers to 6.02e23 atoms so they would have the same amount of atoms
B.
Delta G is negative and Delta S is positive.
The termination step of the free-radical chlorination of methane is the most stable one among all three steps.
The free-radical substitution reaction between chlorine and methane features three major steps:
Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus <em>endothermic</em>.
Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g., 
. The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.
Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.
Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.
Answer:
This is a pretty straightforward example of how an ideal gas law problem looks like.
Your strategy here will be to use the ideal gas law to find the pressure of the gas, but not before making sure that the units given to you match those used by the universal gas constant.
So, the ideal gas law equation looks like this
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
V
=
n
R
T
a
a
∣
∣
−−−−−−−−−−−−−−−
Here you have
P
- the pressure of the gas
V
- the volume it occupies
n
- the number of moles of gas
R
- the universal gas constant, usually given as
0.0821
atm
⋅
L
mol
⋅
K
T
- the absolute temperature of the gas
Take a look at the units given to you for the volume and temperature of the gas and compare them with the ones used in the expression of
R
.
a
a
a
a
a
a
a
a
a
a
a
Need
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Have
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
√
a
a
a
a
a
a
a
Kelvin, K
a
a
a
a
a
a
a
a
a
a
a
a
Celsius,
∘
C
a
a
a
a
a
a
a
a
a
×
Notice that the temperature of the gas must be expressed in Kelvin in order to work, so make sure that you convert it before plugging it into the ideal gas law equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
T
[
K
]
=
t
[
∘
C
]
+
273.15
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Rearrange the ideal gas law equation to solve for
P
P
V
=
n
R
T
⇒
P
=
n
R
T
V
Plug in your values to find
P
=
0.325
moles
⋅
0.0821
atm
⋅
L
mol
⋅
K
⋅
(
35
+
273.15
)
K
4.08
L
P
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
2.0 atm
a
a
∣
∣
−−−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the temperature of the gas.
