An ore contains 12% copper. How many kilograms of the ore are required to get 69 kg
of copper?

Answer is: 315,29 grams of copper ore.

m(chalcopyrite) = ?

ω(CuFeS₂) = 55,0 % = 0,55.

m(Cu) = 60,0 g.

mass percentage of copper in CuFeS₂:

ω(Cu) = Ar(Cu) ÷ Mr(CuFeS₂).

ω(Cu) = 63,55 ÷ 183,4.

ω(Cu) = 0,346 = 34,6 %.

mass percentage of copper in chalcopyrite:

ω(Cu) : ω₁(Cu) = 100% : ω(CuFeS₂).

ω₁(Cu) = 19,03 % = 0,1903.

m(chalcopyrite) = 60,0 g ÷ 0,1903.

m(chalcopyrite) = 315,29 g.

Step-by-step explanation: Im sorry if thats alot o.o

575 kg

Step-by-step explanation:

100 kg of ore = 12 kg of copper

69kg of copper = x kg of ore

12x = 100 x 69

x = 6900 / 12

x = 575

560

hope this : )

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c