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2 years ago

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There are two steps in the extraction of copper metal from chalcocite, a copper ore. In the first step, copper(I) sulfide and ox

ygen react to form copper(I) oxide and sulfur dioxide:
2Cu2S+ 3O2 â 2Cu2O + 2SO2

In the second step, copper(I) oxide and carbon react to form copper and carbon monoxide:

Cu2 O(s)+ C(s) â 2Cu(s)+ CO(g)

Write the net chemical equation for the production of copper from copper(I) sulfide, oxygen and carbon.

1 answer:

maxonik [38]2 years ago

4 0

Answer:

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Explanation:

Equation 1 should correctly be written as;

2Cu2S + 3O2-----> 2Cu2O + 2SO2

Equation 2 should be correctly written as;

2Cu2O + 2C -----> 4Cu + 2CO

The overall reaction equation is;

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Note that species that are intermediates are cancelled out .

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What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a

Blababa [14]

First, we write the balanced equation for this reaction:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

Moles = volume (in L) * molarity

We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L

The volume required is 173.6 mL

5 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

What does the second law of thermodynamics say about entropy

alukav5142 [94]

Answer:

The entropy and the systems surrounding it tend to increase.

5 0

2 years ago

Read 2 more answers

Law Incorporation [45]

Answer:

They are magnetic lines and theses lines depict the invisible magnetic field around the magnet! Hope this helps I got it right on my quiz:)

Explanation:

3 0

3 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

3 0

3 years ago