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Oduvanchick [21]
3 years ago
7

A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials an

d waste disposal) and the earnings (the price at which the product can be sold). calculate the profit from producing 62.00 kg of propene oxide.
Chemistry
1 answer:
Maksim231197 [3]3 years ago
7 0
<span>Answer: it shows that 1mol mCPHA provides the oxygens to 1 mol of propene, to make 1 mole of C3H6O so: 1 mol C3H6 & 1 mol mCPHA --> 1 mol C3H6O using molar masses, that equation becomes: 42.08grams C3H6 & 172.57grams mCPHA --> 58.08grams C3H6O which is: 42.08 kg C3H6 & 172.57 kg mCPHA --> 58.08 kg C3H6O to produce 1 kg of C3H6O, this becomes: 42.08 / 58.08 kg C3H6 & 172.57 / 58.08 kg mCPHA --> 58.08 /58.08 kg C3H6O which is: 0.72452 kg C3H6 & 2.9712 kg mCPHA --> 1 kg C3H6O but because the reaction gives only a 96% yield, we scale up the reactants to get that desired 1 kg of C3H6O (0.72452 kg ) (100/96) C3H6 & (2.9712 kg) (100/96) mCPHA --> 1 kg C3H6O which is: 0.75471 kg C3H6 & 3.095 kg mCPHA --> 1 kg C3H6O ========= costs per kg of C3H6O produced: (0.75471 kg C3H6) ($10.97 per kg) = $8.279 (3.095 kg mCPHA) ($5.28 per kg) = $16.342 & (0.75471 kg C3H6) / (0.0210 kg C3H6 / L dichloromethane) = 35.939 Litres dichloromethane (35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19 & waste disposal is $5.00 per kilogram of propene oxide produced total cost, disregarding labor,energy, & facility costs: $8.279 & $16.342 & $ $ 76.19 & $5.00 = $105.81 per kg C3H6O produced ========== profit: ($258.25 / kg C3H6O) - ($105.81 cost per kg) = $152.44 profit /kg “Calculate the profit from producing 75.00kg of propene oxide” (75.00kg) ($152.44 /kg) = $11,433 that answer rounded off to four sig figs, is $11,430</span>
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The electronegativity values of carbon, hydrogen, and nitrogen are compared in the table.
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Explanation:

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Hydrogen bonds are one of the strongest intermolecular forces. It is as a result of the electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom N, O and F of another molecule.

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Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

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Diano4ka-milaya [45]

Answer:

The first one is B, "Decreasing surface area."

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