<u>Answer:</u>
<u>Plasmas of great interest to scientists or manufacturers as</u>
- Plasma is electrically charged gases that contain considerable charged particles that can change the behavior of the substance.
<u>Current uses of plasmas:</u>
- First, it is used to make semiconductors for different types of electronic equipment
- Secondly, they're used in making transmitters for high-temperature films.
<u>Way scientists and engineers hope to use plasmas in the future:</u>
- The scientists are hoping to use plasma in the future to get rid of all hazardous wastes through a process called plasma gasification.
Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,
The boiling point of hydrocarbons generally increases as the size of the molecules increases because more bonds are needs to be broken in larger organic molecules.
<h3>What are hydrocarbons?</h3>
Hydrocarbons are organic compounds which here composed of hydrogen and carbon alone.
Hydrocarbons are grouped into families or homologous series based on a reactive group known as the gincyiial group
The homologous series include
The boiling point generally increases as the size of the molecules increases because more bonds are needs to be broken in larger organic molecules.
Learn more about hydrocarbons at: brainly.com/question/3551546
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Answer: 167 g
Explanation:
1) The depression of the freezing point of a solution is a colligative property ruled by this equation:
ΔTf = i × m × Kf
Where:
ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.
i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1
Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C
2) Calculate the molality (m) of the solution
ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m
3) Calculate the number of moles from the molality definition
m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent
moles of solute = 2.69 m × 1.00 kg = 2.69 moles
4) Convert moles to grams using the molar mass
molar mass of C₂H₆O₂ = 62.07 g/mol
mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g
