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4 years ago

18 E. Cu2O(s)+O2(g)⟶CuO(s)

Chemistry

1 answer:

iren2701 [21]4 years ago

6 0

Answer:

Balance the reaction of Cu2O + O2 = CuO using this chemical equation ... O₂ Oxygen Molecular Oxygen Diatomic Oxygen Triplet Oxygen Oxygen Gas Liquid .

Explanation:

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Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o

Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

$\tt aA+bB\rightarrow cC+dD$

The rate reaction :

$\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}$

Reaction for formation CCl₄ :

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂

Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0

3 years ago

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.