Context:
175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)
The equation that balances this reaction is listed here:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.
Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
This results in 185.94 kg of NH3 required
For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg
To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O
Answer:
4.86×10^23 lead molecules
Explanation:
Using the provided equation, 2 moles of ammonia produces 3 moles of lead.
Thus:
2 mol NH3/ 3 mol Pb
By utilizing this proportion, we can calculate the quantities for each molecule. For 5.38 mol NH3:
(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb
Next, we will apply Avogadro's number to determine the total number of molecules.
(8.07)(6.02×10^23) = 4.86×10^23 lead molecules
In this scenario, the average cofactor mass for each cell is indicated as 77.91pg, and the total cell quantity is given as 10^5 cells. You need to calculate the total cofactor that will be present in those cells (where 1 microgram equals 10^6 picograms). The calculation follows:
Cofactor mass = cofactor per cell * total cell count = 77.91pg/cell * 10^5 cells = 7.791 x 10^6pg
Next, to convert picograms to micrograms: 7.791 x 10^6pg / (10^6pg/microgram) = 7.791 micrograms
Response:
a. The return of the process fluid after it has been cooled, condensed, or heated in a distillation column or packing.
b. should remain stable at elevated temperatures
should not generate additional by-products
should be chemically inert
c. expressed as a range
d. determined by conducting a series of measurements
e. impurities cause substances to melt at lower temperatures.
Explanation:
a. In a reflux system, the condensate is returned to the initial flask or boiler. This keeps the process fluid at a constant temperature without needing more fuel while increasing the molar fraction of the distillate (which enhances product purity).
b. A solvent needs to be stable and should not participate in side reactions that produce unwanted by-products, ideally remaining inert at elevated temperatures.
c. It is advisable to report melting points as a range since various factors influence the melting point in experimental setups.
d. Melting points can be ascertained through a series of graphical measurements.
e. Impurities reduce the melting point of substances.
