"native," or elemental copper can be found in nature, but most copper is mined as oxide or sulfide minerals. chalcopyrite (cufes2) 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 $$2cus cu2s+s $$cu2s+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?

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.

the answer to your question is c

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.