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galben [10]
3 years ago
10

2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti

on, the temperature is 26.55◦C. How much heat is evolved per gram of fuel burned? The heat capacity of the calorimeter (hardware only) is 92.3 J/◦C.
Chemistry
1 answer:
olga2289 [7]3 years ago
7 0

Answer:

\large \boxed{\text{933 J}}

Explanation:

There are three heat transfers involved.

heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0

  q₁     +     q₂      +       q₃      = 0

m₁ΔH + m₂C₂ΔT + C_calΔT = 0

Data:

 m₁ =      2.1   g

 m₂ = 280     g

   Ti = 25.00 °C

   T_f = 26.55 °C

Ccal = 92.3 J·°C⁻¹

Calculations:

Let's calculate the heats separately.

1. q₁

q₁ = 2.1 g × ΔH = 2.1ΔH g

2. q₂

ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C

q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J

3. q₃

q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J

4. ΔH

\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}

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Answer:

x = 1, -7.5

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3 years ago
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Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
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Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

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