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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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3 years ago
Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li than for K
OLEGan [10]

Li+ has a smaller ionic radius than K+

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<h3>What is ion-solvent interaction ?</h3>

In the case of ion-solvent interactions, the state in which the interac-tions exist is an obvious one; it is the situation in which ions are inside the solvent.

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6 0
1 year ago
How many molecules are in 1kg of water
Mila [183]

Answer:

334.2× 10²³ molecules

Explanation:

Given data:

Mass of water = 1 Kg ( 1000 g )

Number of molecules = ?

Solution:

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Number of moles = mass/ molar mass

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Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres
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Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

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Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

Ksp=[Ag^+][Cl^-]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

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I          -                   0             0

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E        -                    x             x

Which leads to the following modified equilibrium expression:

Ksp=x^2

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

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