<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>
Answer:
B i'm pretty sure, correct me if i'm wrong.
Explanation:
As, overall order of reaction is 2 indicates the rate of reaction is only depending on concentration of A and its rate of reaction can be shown as:
R = K [A]^2
So, by increasing the concentration of A, the rate of reaction increases.
Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.
Explanation:
- Firstly, we need to define the term <em>"latent heat"</em> which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
- Types of latent heat: depends on the phases that the change occur between them;
- Liquid → vapor, <em>latent heat of vaporization</em> and energy is absorbed.
- Vapor → liquid, latent heat of liquification and the energy is removed.
- Liquid → solid, <em>latent heat of solidification</em> and the energy is removed.
- Solid → liquid, <em>latent heat of fusion</em> and the energy is absorbed.
- In our problem, we deals with latent heat of freezing "solidification" of water.
- The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
- Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.
