The answer is chemical weathering
<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>
To find the molecular formula from the empirical formula, you need to find a multiple (x) that will give you the molar mass of the compound which in the question is 54 g/mol.
If C₂H₃ is the empirical formula
molar mass of empirical formula = (12 × 2) + (1 × 3) g/mol
= 27 g/mol
let x = multiple
let molecular formula = C₂ₓ H₃ₓ
multiple = molecular mass ÷ empirical mass
= 54 g/mol ÷ 27 g/mol
= 2
If molecular formula = C₂ₓ H₃ₓ
then molecular formula = C₂₍₂₎H₃₍₂₎
= <span>C₄H</span>₆
Answer:
OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20
Explanation:
Butane burns is air (O2) according to the equation:
C4H10 + O2 —> CO2 + H20
Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;
There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:
C4H10 + O2 —> 4CO2 + H20
Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:
C4H10 + O2 —> 4CO2 + 5H20
Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2
as shown below
C4H10 + 13/2O2 —> 4CO2 + 5H20
Now we multiply through by 2 clear off the fraction and we obtained:
2C4H10 + 13O2 —> 8CO2 + 10H20
Answer:
157.29 Torr
Explanation:
This question may be solved by the mole fraction.
Mole fraction → Moles of gas / Total moles = Partial pressure of gas / Total pressure
If the atmosphere is about 21% you assume that the mole fraction is 0.21 so:
0.21 = Partial pressure of gas / Total pressure
0.21 = Partial pressure of gas / 749 Torr
Partial pressure of the gas = 749 Torr . 0.21 → 157.29 Torr