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Romashka [77]
3 years ago
14

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 30.0 g carbon and 40.0 g oxygen. Another sam

ple of ascorbic acid, isolated from lemons (an excellent source of the vitamin), contains 12.7 g carbon. Compute the mass of oxygen (in grams) in the second sample.
Chemistry
1 answer:
mote1985 [20]3 years ago
8 0

Answer:

16.933g approximately 17.0 grams

Explanation:

From the simple promotions and given the same compound ascorbic acid (vitamin C)

In the laboratory synthesised ascorbic acid

Mass of carbon = 30.0g

Mas of Oxygen = 40.0g

That is the mass of Oxygen per unit mass of Carbon

Per gram of Carbon we have

(30.0g Carbon)÷30 combines with (40.0g of Oxygen)÷30

That is 4/3g of Oxygen per gram of Carbon

Hence the mass of Oxygen compound that combines with 12.7g of Carbonin natural occurring ascorbic acid (vitamin C) is = 4/3×12.7 = 16.933g approximately 17.0g

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Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

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