The theory of evolution was proposed by Darwin.
Answer:
Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.
Hope this helped and if it did, please give my answer a brainliest.
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.
1.
Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar
2. For the volume, let's find the total volume first.
V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
#SPJ4
Kepler's
third law shows the relationship between the orbital period of an object and
the distance between the object and the object it orbits.
The
simplified version of this law is: P^2 = a^3
Where,
P =
period of the orbit in years = 0.62 years
a =
average distance from the object to the object it orbits in AU. The
astronomical unit AU is a unit of length which is roughly equivalent to the
distance from Earth to the Sun.
Therefore
calculating for a:
0.62
^ 2 = a ^ 3
a =
0.62 ^ (2/3)
a =
0.727 AU = 0.72 AU
Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to
Sun.
<span>Answer:
B. 0.72 AU</span>
