<span>(8.90/95.211) =0.09347 moles</span>
The answer to the question is 300meters.
Question 1:
(a) Sulfurous acid: H2SO3
Sulfuric acid: H2SO4
(b) Nitrous acid: H2NO2
Nitric acid: H2NO3
Question 2:
To calculate the pH, based on concentration of H+ ions, there is one formula:
So the pH of this solution is
(the solution is basic).
HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.
From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
However, this is the molarity of the HCl and NaOH solution, not the original HCl solution. Using the dilution equation M1V1=M2V2, we can solve for the original molarity.
M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
<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>
