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aalyn [17]
3 years ago
11

"native," or elemental copper can be found in nature, but most copper is mined as oxide or sulfide minerals. chalcopyrite (cufes

2) is one copper mineral that can be converted to elemental copper in a series of chemical steps. reacting chalcopyrite with oxygen at high temperature produces a mixture of copper sulfide and iron oxide. the iron oxide is separated from cus by reaction with sand (sio2). cus is converted to cu2s in the process and the cu2s is burned in air to produce cu and so2: $$2cufes2​+3o2​ 2cus+2feo+2so2​ $$2feo+2sio2​ 2fesio3​ $$2cus cu2​s+s $$cu2​s+s+2o2​ 2cu+2so2​ 3rd attempt part 1 (1 point)feedbacksee hintsee periodic table an average copper penny minted in the 1960s contained about 3.000 g of copper. how much chalcopyrite had be mined to produce 100 pennies? 867.02 g cufes2 part 2 (1 point)feedback how much chalcopyrite had to be mined to produce 100 pennies if reaction 1 had a percent yield of 80.00 % and all other reaction steps had yield of 100%? 1083.75 g cufes2 part 3 (1 point)feedback how much chalcopyrite had be mined to produce 100 pennies if each reaction involving copper proceeded in 80.00 % yield?
Chemistry
2 answers:
max2010maxim [7]3 years ago
7 0
I acually habe no idea but I nust want to point out that my toast's brand is Sarah Lee lol
neonofarm [45]3 years ago
5 0

1) Balanced chemical reaction 1: 2CuFeS₂ + 3O₂ → 2CuS + 2FeO + 2SO₂.  

Balanced chemical reaction 2: 2FeO + SiO₂ → 2FeSiO₃.  

Balanced chemical reaction 3: 2CuS → Cu₂S + S.

Balanced chemical reaction 4: Cu₂S + S + O₂ → 2Cu + 2SO₂.

m(Cu) = 3.0 g; mass of copper in one penny.

n(Cu) = m(Cu) ÷ M(Cu).

n(Cu) = 3 g ÷ 63.55 g/mol.

n(Cu) = 0.047 mol; amount of copper in one penny.

2) From chemical reaction 4: n(Cu) : n(Cu₂S) = 2 : 1.

n(Cu₂S) = 0.047 mol ÷ 2.

n(Cu₂S) = 0.0236 mol; amount of copper(I) sulfide.

From chemical reaction 3: n(Cu₂S) : n(CuS) = 1 : 2.

n(CuS) = n(Cu₂S) · 2.

n(CuS) = 0.047 mol; amount of copper(II) sulfide.

From balanced reaction 1: n(CuS) : n(CuFeS₂) = 2 : 2 (1 : 1).

n(CuFeS₂) = 0.047 mol; amount of chalcopyrite.

m(CuFeS₂) = n(CuFeS₂) · M(CuFeS₂).

m(CuFeS₂) = 0.047 mol · 183.54 g/mol.

m(CuFeS₂) = 8.67 g; mass for one penny.

m(CuFeS₂) = 8.66 g · 100 = 867.02 g; for 100 pennies.

3) From balanced chemical reactions: n(Cu) : n(CuFeS₂) = 1 : 1.

n(CuFeS₂) = 0.047 mol.

ω(CuFeS₂) = 80.00% ÷ 100%.

ω(CuFeS₂) = 0.800; a percent yield of chemical reaction 1.

m(CuFeS₂) = 0.047 mol · 183.54 g/mol ÷ 0.800.

m(CuFeS₂) = 10.83 g; mass of chalcopyrite for one penny.

m(CuFeS₂) = 10.83 g · 100.

m(CuFeS₂) = 1083.75 g; mass of chalcopyrite for 100 pennies.

4) From chemical reaction 4: n(Cu) : n(Cu₂S) = 2 : 1.

n(Cu₂S) = 0.047 mol ÷ 2.

n(Cu₂S) = 0.0236 mol ÷ 0.8.

n(Cu₂S) = 0.0295 mol; amount of copper(I) sulfide with 80.0% yield.

From chemical reaction 3: n(Cu₂S) : n(CuS) = 1 : 2.

n(CuS) = n(Cu₂S) · 2 ÷ 0.80.

n(CuS) = 0.07375 mol; amount of copper(II) sulfide with 80.0% yield.

From balanced reaction 1: n(CuS) : n(CuFeS₂) = 2 : 2 (1 : 1).

n(CuFeS₂) = 0.07375 ÷ 0.80.

n(CuFeS₂) = 0.0922 mol; amount of chalcopyrite with 80.0% yield.

m(CuFeS₂) = n(CuFeS₂) · M(CuFeS₂).

m(CuFeS₂) = 0.0922 mol · 183.54 g/mol.

m(CuFeS₂) = 16.92 g; mass for one penny.

m(CuFeS₂) = 16.92 g · 100 = 1692.0 g; for 100 pennies.

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How many moles of CaO will be produced from 23.9 g of Ca?<br>2Ca(s) + O2(g) – 20:0(3)<br>mol​
inessss [21]

Answer:

0.6 moles of CaO will produced.

Explanation:

Given data:

Mass of calcium = 23.9 g

Moles of CaO produced = ?

Solution:

Chemical equation:

2Ca + O₂ → 2CaO

Number of moles of calcium:

Number of moles = mass/ molar mass

Number of moles = 23.9 g / 40 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of calcium and CaO.

            Ca         :          CaO

              2          :          2

            0.6         :        0.6

 0.6 moles of CaO will produced.

4 0
3 years ago
After an electric sign is turned on, the temperature of its glass goes from 23.5°C to 65.5°C. The sign’s glass has a mass of 905
julia-pushkina [17]

The sign's glass absorbed 25466.7 J

<h3>Further explanation</h3>

Given

The temperature of glass :  23.5 °C to 65.5 °C

mass = 905 g

the specific heat capacity = 0.67 J/g °C

Required

Heat absorbed

Solution

Heat absorbed by sign's glass can be formulated :

\tt Q=m.c.\Delta T

ΔT=65.5 - 23.5 = 42

\tt Q=905\times 0.67\times 42\\\\Q=\boxed{\bold{25466.7~J}}

5 0
3 years ago
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Covalent and hydrogen bonds

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
marysya [2.9K]

Answer:

\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) +   e⁻ ⇌ Ag(s)     E° =  0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)     E° = -0.447    V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

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