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.
Answer:
A. move materials through the body
Explanation:
The blood circulatory system (cardiovascular system) transports materials and delivers nutrients and oxygen to all cells in the body.
SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
First, we will calculate the partial pressure of SF₆ using the following expression.
where,
- pSF₆: partial pressure of SF₆
- P: total pressure of air (we will assume it is 1 atm)
- ppb: concentration of SF₆ in parts per billion
Then, we will convert 1.0 cm³ to L using the following conversion factors:
- 1 cm³ = 1 mL
- 1 L = 1000 mL
Next, we will convert 46 °C to Kelvin using the following expression.
Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.
Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.
SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
You can learn more about partial pressure here: brainly.com/question/13199169
By gaining an electron (perhaps via beta particle). The H+ ion needs another electron to remove its positive charge.
