Respuesta:
72 L
Explicación:
Paso 1: Calcular los moles de CO
La masa molar de CO es 28,01 g/mol.
Paso 2: Convertir la temperatura a Kelvin
Siempre que trabajamos con gases ideales, debemos convertir las temperaturas a la escala Kelvin. Usaremos la siguiente formula.
K = °C + 273,15
K = 27°C + 273.15
K = 300 K
Paso 3: Calcular el volumen de gas
Usaremos la ecuación del gas ideal
A which is element is the answer
Answer:
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Answer:
Methyl Alcohol.
Acetate
Heptane
Dioxane
Explanation:
The boiling point of liquids are measure by observing the gas pressure or vapors temperature. There are few liquids whose boiling point is difficult to measure. The gas pressure differs for every liquid. It is important to observe the vapor pressure also to identify the boiling point. The most difficult boiling point identification is for Methyl alcohol as it has ability to absorb different level of pressure.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
