Determine the Concentration of the Unknown Strong Acid

Question 9 (1 point)

pantera1 [17]

Answer:

8 moles of C

Explanation:

From the question given above, the following equation was obtained:

3A + 2B —> 6C

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:

From the equation above,

3 moles of A reacted to produce 6 moles of C.

Therefore, 4 moles of C will react to produce = (4 × 6)/3 = 8 moles of C

Thus, 8 moles of C can be obtained from the reaction of 4 moles of A with excess B

6 0

3 years ago

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

hjlf

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.

4 0

3 years ago

According to the kinetic molecular theory particles of matter

DiKsa [7]

Answer:

The kinetic molecular theory of matter states that: Matter is made up of particles that are constantly moving. ... Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.

Explanation:

give me brainliest please

4 0

3 years ago

A baseball has a mass of 135 grams and a softball has a mass of 270 grams. In which of the following situations would they have

liraira [26]

Answer:

The baseball is thrown twice as fast as the softball in the same direction.

Explanation:

8 0

3 years ago

An atom of an element that has the same number of protons but a different number of neutrons is called a(n): ion nucleus isotope

alexandr1967 [171]

It is called an isotope

5 0

3 years ago