What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C
2 answers:
Answer:
4.704625 kJ (or 4704.625 J) is the heat obtained when 75 grams of water's temperature rises from 20°C to 35°C.
Explanation:
The process of quantifying heat transfer between systems or bodies is termed calorimetry.
In this context, there is a linear relationship between heat and temperature change. The proportionality constant relies on the substance's properties, derived from the mass times the specific heat. The specific equation utilized for calculating heat transfer is:
Q = c * m * T
Where Q denotes the heat exchange for a mass m, with a specific heat c, and T indicating the temperature difference.
In this scenario:
- The specific heat of water is 4,181 J/g°C
- m=75 g
- T=35°C - 20°C = 15°C
Thus:
Q≅4704.625 J
Since 1 kJ = 1000 J:
Q=4.704625 kJ
Ultimately, 4.704625 kJ (or 4704.625 J) reflects the heat absorbed when 75 grams of water's temperature shifts from 20°C to 35°C
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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: The rate constant at 525 K is,
Rationale:
Based on the Arrhenius equation,
or,
where,
= rate constant when
=
= rate constant when
=?
= activation energy for the process =
R = gas constant = 8.314 J/mole.K
= initial temperature = 701 K
= final temperature = 525 K
Substituting the provided values into this formula yields:
Thus, the rate constant at 525 K is,