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:

The question is incomplete, the correct question is:

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?

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

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.