Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Answer:
Explanation:
1. Mass of zinc
The density of zinc is 7.14 g/cm³
D = m/V
2. Volume of aluminium
The density of aluminium is 2.60/cm³.
Answer:
C. 548mL
Explanation:
Attached is an image of the explanation. Hope this helps!
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
.
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Have
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Liters, L
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Liters, L
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√
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Kelvin, K
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Celsius,
∘
C
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×
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.
