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2 answers:
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Answer:
1. 4-ethyl-1-heptene
2. 6-ethyl-2-octene
3. 1-butyne
Explanation:
The compounds are named according to IUPAC rules.
Compound 1:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 7 carbon atoms, so taken the name hept.
- The double bond between C1 and C2, so take no. 1 and add the suffix ene to hept "1-heptene".
- The ethyl group is the alkyl substituent on position 4.
- So the name is 4-ethyl-1-heptene.
Compound 2:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 8 carbon atoms, so taken the name oct.
- The double bond between C2 and C3, so take no. 2 and add the suffix ene to oct "2-octene".
- The ethyl group is the alkyl substituent on position 6.
- So the name is 6-ethyl-2-octene.
Compound 3:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain), there is no substituents.
- The parent chain is numbered so that the multiple bonds have the lowest numbers (Triple bond here take the lowest number).
- The longest chain contains 4 carbon atoms, so taken the name but.
- The triple bond between C1 and C2, so take no. 1 and add the suffix yne to but "1-butyne".
- So the name is 1-butyne.
Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,
<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:
P₂ in terms of the ideal gas equation is: