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

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

h/o/e/s mad

step-by-step explanation:

si si si