A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes
1 answer:
0.208 is the specific heat capacity of the metal.
Explanation:
Given:
mass (m) = 63.5 grams 0R 0.0635 kg
Heat absorbed (q) = 355 Joules
Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K
cp (specific heat capacity) = ?
the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:
q = mc Δ T
c =
c =
= 0.208 J/gm K
specific heat capacity of 0.208 J/gm K
The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.
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Answer:
P IBr: 15.454atm
I₂: 0.923 atm
P Br₂: 0.923atm
Explanation:
Basados en la reacción:
I₂(g) + Br₂(g) ⇄ 2 IBr(g)
La constante de equilibrio, Keq, es definida como:
<em>Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio</em>
Usando PV = nRT, la presión inicial de IBr es:
P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = <em>17.3 atm</em>
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Siendo las presiones en equilibrio:
P IBr: 17.3 - 2X
P I₂: X
P Br₂: X
<em>Donde X representa el avance de reacción.</em>
Remplazando en Keq:
280 = (17.3 - 2X)² / X²
280X² = 4X² - 69.2X + 299.29
0 = -276X² - 69.2X + 299.29
<em>Resolviendo para X:</em>
X = -1.174 → Solución falsa. No existen presiones negativas
X = 0.923 → Solución real
Así, las presiones parciales en equilibrio de cada compuesto son:
P IBr: 17.3 - 2X = <em>15.454atm</em>
P I₂: X =<em> 0.923atm</em>
P Br₂: X = <em>0.923atm</em>