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Nimfa-mama [501]
2 years ago
13

A metallurgist has one alloy containing 21% aluminum and another containing 42% aluminum. how many pounds of each alloy must he

use to make 41 pounds of a third alloy containing 38% aluminum?
Chemistry
1 answer:
Naily [24]2 years ago
8 0
Answer is: 7.8 lb of 21% aluminum and 33.2 ib of <span>42% aluminum.</span>

ω₁<span> = 21% ÷ 100% = 0.21.
ω</span>₂<span> = 42% ÷ 100% = 0.42.
ω</span>₃<span> = 38% ÷ 100% = 0.38.
</span>m₁ = ?.

m₂<span> = ?.
</span>m₃ = m₁ + m₂<span>.
</span>m₃ = 41 pounds.

m₁ = 41 lb - m₂<span>.
ω</span>₁ · m₁ + ω₂ ·m₂ = ω₃ · m₃.

0.21 · (41 lb - m₂) + 0.42 · m₂ = 0.38 · 41 lb.

8.61 lb - 0.21m₂ + 0.42m₂ = 15.58 lb.

0.21m₂ = 6.97 lb.

m₂ = 6.97 lb ÷ 0.21.

m₂ = 33.2 lb.

m₁ = 41 lb - 33.2 lb.

m₁<span> = 7.8 lb.</span>



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