Draw the structural formula of 4,5-diisopropylnonane.
1 answer:
According to IUPAC guidelines, the primary carbon chain should have 9 carbon atoms, as indicated by the term 'nonane'. From this chain, two isopropyl groups (each with 3 carbon atoms) should be connected to the 4th and 5th carbons, starting from either end of the main carbon chain. The structural representation is displayed in the attached image.
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Answer:
Explanation:
Diethyl malonate possesses greater acidity compared to monocarbonyl substances (pKa=13) because its alpha hydrogens are linked to two carbonyl groups. Consequently, the malonic ester can be readily changed into its enolate ion by reacting it with sodium ethoxide in ethanol. When the malonic ester undergoes alkylation, a hydrogen atom in the alpha position becomes acidic, permitting another round of alkylation to yield a dialkylated malonic ester.
In this scenario, when diethyl malonate interacts with urea in the presence of sodium ethoxide base, the second alkylation step occurs within the molecule, producing a cyclic compound known as barbituric acid.
Respuesta:
0.16 M
Explicación:
Teniendo en cuenta:
O sea,
Dado que:
Para :
Molaridad = 0.200 M
Volumen = 20.0 mL
Convierte mL a L:
1 mL = 10⁻³ L
Entonces, volumen = 20.0×10⁻³ L
Los moles de son:
Moles de = 0.004 moles
Para :
Molaridad = 0.400 M
Volumen = 30.0 mL
Convertimos mL a L:
1 mL = 10⁻³ L
Volumen = 30.0×10⁻³ L
Entonces, los moles de son:
Moles de = 0.012 moles
Según la reacción:
1 mol de reacciona con 1 mol de
Por lo tanto,
0.012 mol de reacciona con 0.012 mol de
Moles disponibles de = 0.004 mol
El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).
La formación del producto depende del reactivo limitante, así que,
1 mol de reacciona con 1 mol de
y produce 1 mol de
0.004 mol de reacciona con 0.004 mol de
y genera 0.004 mol de
Los moles restantes de son: 0.012 - 0.004 = 0.008 mol
El volumen total es 20 + 30 mL = 50 mL = 0.050 L
Por lo que la concentración del ion bario, , después de la reacción es:
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.
