What is the term for the amount of space occupied by an object?

Copper has been used for thousands of years, either as a pure metal or in alloys. It is frequently used today in the production of wires and cables. Copper can be obtained through smelting or recycling. Determine the energy associated with each of these processes in order to recycle 1.08 mol Cu. The smelting of copper occurs by the balanced chemical equation: CuO(s)+CO(g) Cu(s)+CO2?(g) where ?H°f, CuO is = -155 kJ/mol. Assume the process of recycling copper is simplified to just the melting of the solid Cu starting at 25°C. The melting point of Cu is 1084.5°C with ?H°fus = 13.0 kJ/mol and a molar heat capacity, cp,Cu = 24.5 J/mol·°C.

Enthalpy change for the reaction recovering ?

Cu from CuO Energy for recycling Cu?

Answer:

Energy for recovering Cu from CuO = - 138. 24kJ

the total energy for recycling Cu is 42.07kJ

Explanation:

CuO(s) + CO(g) - - - - - - - > Cu(s) + CO2(g)

ΔHrxn = ΔHf(products) - ΔHf(reactants)

= ΔHf(CO2) - (ΔHf(CO)) + Δ Hf(CuO))

= - 393.5 kJ/mol - (-110.5 kJ/mol + ( - 155 kJ/mol)

= - 393.5 kJ/mol + 265.5 kJ/mol

= - 128 kJ/mol

for 1.08 mol of Cu

ΔH= - 128 kJ/mol × 1.08 mol = - 138. 24 kJ

Therefore,

Energy for recovering Cu from CuO = - 138. 24kJ

Part.2 :-

Total energy required = Heat required to raise the temperature of Cu from 25°C to 1084.5°C (q1) + Heat required to melt Cu at 1084.5°C(q2)

q1= n × ΔT × Cp

q1 = 1.08 mol × (1084.5°C - 25°C) × 24.5 J/mol 0C

q1 = 28.03 kJ

q2 = ΔHfus × n

q2 = 13.0 kJ/mol × 1.08 mol

q2 = 14.04kJ

Therefore,

Energy for recycling Cu = 28.03 kJ + 14.04kJ = 42.07kJ

Therefore, the total energy for recycling Cu is 42.07kJ

4 0

3 years ago

Please give explanation as well

jekas [21]

Answer:

Concentration of HCl = $\frac{0.01}{0.134}$ = 0.075L= 75mL

Concentration of KOH = $\frac{0.002}{0.357}$ = 0.0056L = 5.6mL

Concentration of H₂SO₄ = $\frac{0.014}{0.13}$ = 0.108L = 108mL

Explanation:

Procedure

This a problem related to volumetric analysis and we need to find the concentrations of the acids involved in the neurtralization process.

In order to determine the concentrations, we work from the known reactants to the unknown reactants or products.

1. Write the balanced equation of the reactions

2. List the given parameters and work from the known to the unknown. The known is that specie that can give us the number of moles required for this reaction.

3. Check the parameters and make sure that they are in their appropriate units.

4. Obtain the number of moles of the known using the concentration and volume of the reactant using the equation below:

Number of moles = Concentration x Volume.

5. Using the known number of moles, determine that of the unknown by comparing their mole ratios.

6. Since we have obtained the number of moles of the unknown, we can then solve for the concentration of the unknown using the expression below:

$\frac{number of moles}{volume}$

Solution

1. If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

Given parameters:

Volume of base NaOH = 67mL to litres = 67 x 10⁻³ = 0.067L

Concentration of NaOH = 0.15M

Volume of acid HCl = 134mL = 134 x 10⁻³ = 0.134L

Concentration of acid = ?

Equation of reaction: NaOH + HCl → NaCl + H₂O

The known here is the base, NaOH.

Using:

Number of moles of NaOH = Concentration of NaOH x Volume of NaOH

Number of moles of NaOH = 0.15M x 0.067L = 0.01mol

From the equation of the reaction, we know that;

1mole of NaOH reacted with 1mole of HCl

Therefore, 0.01 mole of HCl would also react with 0.01 of NaOH

Now that we know the number of moles HCl, we can now obtain the concentration of HCl required to neutralize NaOH using the equation below:

Concentration of HCl = $\frac{number of moles}{volume}$

Concentration of HCl = $\frac{0.01}{0.134}$ = 0.075L= 75mL

2. If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

Given parameters:

Concentration of H₂SO₄ = 0.05M

Volume of acid H₂SO₄ = 27.4mL = 27.4 x 10⁻³ = 0.0274L

Volume of base KOH = 357mL to litres = 357 x 10⁻³ = 0.357L

Concentration of KOH = ?

Equation of reaction: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

The known here is the acid, H₂SO₄

Using:

Number of moles of H₂SO₄ = Concentration of H₂SO₄ x Volume of H₂SO₄

Number of moles of H₂SO₄ = 0.05M x 0.0274L = 0.001mol

From the equation of the reaction, we know that;

2moles of KOH reacted with 1mole of H₂SO₄

Therefore, 0.001 mole of H₂SO₄ would also react with 0.002 of KOH

Now that we know the number of moles KOH, we can now obtain the concentration of KOH required to neutralize H₂SO₄ using the equation below:

Concentration of KOH = $\frac{number of moles}{volume}$

Concentration of KOH = $\frac{0.002}{0.357}$ = 0.0056L = 5.6mL

3. If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution H₂SO₄, what is the concentration of the H₂SO₄ solution?

Given parameters:

Volume of base NaOH = 55mL to litres = 55 x 10⁻³ = 0.055L

Concentration of NaOH = 0.5M

Concentration of H₂SO₄ = ?

Volume of acid H₂SO₄ = 130mL = 130 x 10⁻³ = 0.13L

Equation of reaction: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

The known here is the base, NaOH

Using:

Number of moles of NaOH = Concentration of NaOH x Volume of NaOH

Number of moles of NaOH = 0.5M x 0.055L = 0.028mol

From the equation of the reaction, we know that;

2moles of NaOH reacted with 1mole of H₂SO₄

Therefore, 0.028 mole of NaOH would also react with $\frac{1}{2}$ of 0.028, 0.014mole of H₂SO₄

Now that we know the number of moles H₂SO₄, we can now obtain the concentration of H₂SO₄ required to neutralize NaOH using the equation below:

Concentration of H₂SO₄ = $\frac{number of moles}{volume}$

Concentration of H₂SO₄ = $\frac{0.014}{0.13}$ = 0.108L = 108mL

5 0

4 years ago

What conditions lead to the formation of large crystals in an igneous rock?

Bingel [31]

Slowly cooling magma

3 0

3 years ago