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yanalaym [24]
2 years ago
9

If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co

nstant pressure at thesame temperature
Chemistry
1 answer:
vladimir2022 [97]2 years ago
4 0

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g)  → 12CO₂ (g) + 6H₂O (l)

\Delta n_g = (12 - 15)/2 = -3/2

We have;

\Delta H = \Delta U + \Delta n_g\cdot R\cdot T

Where R and T are constant, and ΔU is given we can write the relationship as follows;

H = U + \Delta n_g\cdot R\cdot T

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

\Delta n_g = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

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A balance reads an object A to be 45.1 kg. The balance reads another object B to be 33.46 kg. What is the total weight of both o
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Explanation:

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7 0
3 years ago
Calcula la masa atómica del Hierro y las partículas subatómicas de cada uno de sus isótopos. Fe-54 (5.82%), Fe-56 (91.66%), Fe-5
tino4ka555 [31]

Answer:

La masa atómica del hierro es 55.847 gramos por mol.

Explanation:

Las masas molares de Fe_{54}, Fe_{56}, Fe_{57} y Fe_{58} son 53.940 gramos por mol, 55.935 gramos por mol, 56.935 gramos por mol y 57.933 gramos por mol, respectivamente. La masa atómica del hierro se determina mediante el siguiente promedio ponderado:

M_{Fe} = \frac{5.82}{100}\times \left(53.940\,\frac{g}{mol} \right)+\frac{91.66}{100}\times (55.935\,\frac{g}{mol})+\frac{2.19}{100}\times \left(56.935\,\frac{g}{mol} \right)+\frac{0.33}{100}\times \left(57.933\,\frac{g}{mol} \right)

M_{Fe} = 55.847\,\frac{g}{mol}

La masa atómica del hierro es 55.847 gramos por mol.

8 0
3 years ago
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