If the standard free energy change for the conversion of fructopyranose to fructofuranose is 1.7 kJ/mol, what fraction of the to
1 answer:
The connection between Gibb's free energy and temperature can be described as follows. The value for 
is specified as 1.7 kJ/mol. Additionally, we know that k = ![\frac{[product]}{[substrate]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproduct%5D%7D%7B%5Bsubstrate%5D%7D)
= . Since k has a value of 0.50357 at a temperature of 298 K, we can conclude that
= 0.50357, leading to
+ 1 = 1.50357, which equals
= 0.665. Therefore, we can deduce that 0.665 of the total fructose in the solution exists as fructopyranose.
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The answer is - 0.138 M. The buffer pH can be determined using the Henderson equation. Here, 
acts as a weak acid and 
serves as its corresponding conjugate base. The weak acid has two protons, while the base contains one. The equation can therefore be expressed in terms of protons transferred. Phosphoric acid can donate protons in three stages; the equation we’ve referenced pertains to the second stage, as the acid then has only two protons available and the base only one. Given the concentration of the acid as 0.10 M, we need to calculate the concentration of the base necessary to form a buffer with a pH of exactly 7.0. Substituting the values into the equation leads us to the solution. Cross-multiplying, we find that [base] = 1.38(0.10), yielding [base] = 0.138. Therefore, the concentration of the base needed for the buffer is 0.138 M.
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: