Answer:
answer number C is the correct answer for this
To solve this, let's assume ideal gas behavior.
PV=nRT
Let's solve for n. Convert units to SI units first.
Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K
n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol
The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
=21 torr/min
The balanced chemical reaction is,

The rate of disappearance of
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of
= ![-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
The rate of formation of
= ![\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
As we know that,
=21 torr/min
So,
![-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2021torr%2Fmin%3D10.5torr%2Fmin)
And,
![\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D)
![\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%3D21torr%2Fmin)
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
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
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a
a
a
a
a
Need
a
a
a
a
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a
Have
a
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Liters, L
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a
Liters, L
a
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√
a
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a
Kelvin, K
a
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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.
Answer:
5 mL
Explanation:
As this is a problem regarding <em>dilutions</em>, we can solve it using the following formula:
Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:
We <u>input the data</u>:
- 5.0 M * V₁ = 0.25 M * 100 mL
And <u>solve for V₁</u>: