Answer:
2-methyl-2-butanol and 2-ethoxy-2-methylbutane
Explanation:
When the elimination occurs, a good leaving group must leave the compound and will be replaced. A good leaving group is a nucleophile, which intends to bond with compounds with positive charges. In this case, the bromine and iodine ions are nucleophiles that will leave the compounds.
In the mixture ethanol and water, two compounds must replace the leaving groups, which are strong nucleophiles: OH⁻ and ⁻OCH₂CH₃ (O is a strong nucleophile, and after the bonding, H⁺ will be lost in the ethanol molecule).
So, the two possible products are 2-methyl-2-butanol and 2-ethoxy-2-methylbutane, as shown below.
Answer:
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Explanation:
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Answer:
V₂ = 530.5 mL
Explanation:
Given data:
Initial temperature = 20.0°C
Final temperature = 40.0 °C
Final volume = 585 mL
Initial volume = ?
Solution:
Initial temperature = 20.0°C (20+273 = 293 K)
Final temperature = 40.0 °C (40+273 = 323 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 585 mL × 293 K / 323 K
V₂ = 171405 mL.K / 323 K
V₂ = 530.5 mL
Answer:
See explaination
Explanation:
Balanced net ionic reaction.
Pb(s) + Cl2(g) ----- Pb2+ + 2Cl-
oxidation reaction :Pb(s) --- Pb2+ + 2e-
Reduction reaction:Cl2 + 2e- -----Cl-