Answer:
The scheme most often used currently divides all living organisms into five kingdoms: Monera (bacteria), Protista, Fungi, Plantae, and Animalia. ... The Prokaryotae are now divided into two domains, the Bacteria and the Archaea, as different from each other as either is from the Eukaryota, or eukaryotes.
1) As for its chemical composition, coal is a mixture of high-molecular-weight polycyclic aromatic compounds, such as benzene C6H6, toluene C6H5CH3, xylene C6H4(CH3)2, naphthalene C10H8, anthracene C14H10, pyrene C16H10 and their derivatives with high mass fraction of carbon, as well as of water and volatile substances.
2) The coal asphaltenes have a relatively narrow MWD (full width ≈ 150 amu) with an average molecular weight of ≈340 amu. The petroleum asphaltenes display a broader MWD (full width ≈ 300 amu) and are heavier on average (≈680 amu).
~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )
We tend to feel more responsible when fewer people are around to help out. This is because you feel the need to help since you cannot rely on many others to do it.
Answer: Responsible
Note: Answer above is incorrect!!!
He is Lord; live
Answer:
166 torr
Explanation:
Let’s call ethane Component 1 and propane Component 2.
According to Raoult’s Law,
![p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%20%5Cchi_%7B1%7Dp_%7B1%7D%5E%7B%5Ccirc%7D%5C%5Cp_%7B2%7D%20%3D%20%5Cchi_%7B2%7Dp_%7B2%7D%5E%7B%5Ccirc%7D)
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Data:
p₁° = 304 torr
p₂° = 27 torr
n₁ = n₂
1. Calculate the mole fraction of each component
χ₁ = n₁/(n₁ + n₂)
χ₁ = n₁/n₁ + n₁)
χ₁ = n₁/(2n₁)
χ₁ = ½
χ₁ = 0.0.5
χ₂ = 1- χ₁ = 1- 0.5 = 0.5
2. Calculate the vapour pressure of the mixture
![p_{1} = 0.5 \times \text{304 torr} = \text{ 152 torr}\\p_{2} = 0.5 \times \text{27 torr} = \text{ 13.5 torr}\\p_{\text{tot}} = p_{1} + p_{2} = \text{152 torr + 13.5 torr} = \textbf{166 torr}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%200.5%20%5Ctimes%20%5Ctext%7B304%20torr%7D%20%3D%20%5Ctext%7B%20152%20torr%7D%5C%5Cp_%7B2%7D%20%3D%200.5%20%5Ctimes%20%5Ctext%7B27%20torr%7D%20%3D%20%5Ctext%7B%2013.5%20torr%7D%5C%5Cp_%7B%5Ctext%7Btot%7D%7D%20%3D%20p_%7B1%7D%20%2B%20p_%7B2%7D%20%3D%20%5Ctext%7B152%20torr%20%2B%2013.5%20torr%7D%20%3D%20%5Ctextbf%7B166%20torr%7D)
![\text{The vapour pressure above the solution is $\boxed{\textbf{166 torr}}$}](https://tex.z-dn.net/?f=%5Ctext%7BThe%20vapour%20pressure%20above%20the%20solution%20is%20%24%5Cboxed%7B%5Ctextbf%7B166%20torr%7D%7D%24%7D)