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".
Answer:
A. Zodiac
B. Palingenesis
C. Palabra mysteria
D. Decknamen
The correct answer is D. Decknamen.
Explanation:
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
For this system, we use Dalton's law of partial pressures where the total pressure of a gas mixture is said to be equal to the sum of the partial pressures of the gases. The partial pressure of each gas would be calculated by the product of the mole fraction and the original pressure of the gas. We do as follows:
Total pressure = x1P1 + x2P2
Total pressure = (2.0 / 7.0 )(3.5) + (1.5/7.0)(2.6)
Total pressure = 1.56 atm