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notsponge [240]
3 years ago
7

The cost, C C, of producing x x Totally Cool Coolers is modeled by the equation C = 0.005 x 2 − 0.25 x + 12 C=0.005x2-0.25x+12 H

ow many coolers need to be produced and sold in order to minimize the cost? (Round to the nearest whole number.)
Chemistry
1 answer:
neonofarm [45]3 years ago
3 0

Answer:

25 coolers are need to be produce and sell in order to minimize the cost.

Explanation:

C = 0.005x^2-0.25x+12 ..[1]

Differentiating the given expression with respect to dx.

\frac{dC}{dx}=\frac{d(0.005x^2-0.25x+12)}{dx}

\frac{dC}{dx}=0.01x-0.25+0

Putting ,\frac{dC}{dx}=0

0=0.01x-0.25+0

0.01x=0.25

x = 25

Taking second derivative of expression [1]

\frac{d^2C}{dx^2}=\frac{d(0.01x-0.25)}{dx}=0.01

\frac{d^2C}{dx^2}>0 (minima)

25 coolers are need to be produce and sell in order to minimize the cost.

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Veronika [31]

Answer:

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

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Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

8 0
3 years ago
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How many grams of butane were in 1. 000 atm of gas at room temperature?
dimaraw [331]

The mass in grams of butane at standard room temperature is 53.21 grams.

<h3>How can we determine the mass of an organic substance at room temperature?</h3>

The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:

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1 × 22.4 L = n × (0.0821 atm*L/mol*K×  298 K)

n = 22.4/24.4658 moles

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mass of butane = 0.91556 moles × 58.12 g/mole

mass of butane = 53.21 grams

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6 0
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