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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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Explanation:

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3 years ago
how did the boiling point of plain water compare to that of water with salt? Compared to water with sugar?
elixir [45]

The boiling point of plain water is less than the boiling point of both salt and sugar water.

<h3>What is boiling point?</h3>

Boiling point can be defined as the point when the pressure exerted by the surroundings upon a liquid is equal to the pressure exerted by the vapour of the liquid.

The boiling point of plain water is 100°C which increases upon addition of solute substances such as salt and sugar.

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2 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
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anygoal [31]

From the equation,

4 mole of lithium produces 2 mole of lithium oxide

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5 0
3 years ago
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Yes

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