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In-s [12.5K]
3 years ago
14

During the formation of an ionic compound, the element with the higher ionization energy will gain electrons.

Chemistry
1 answer:
Lesechka [4]3 years ago
8 0
False because you have to take it out and do it right
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Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr
kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm    5.1 atm    0       Initial

-2x             -x           +2x    Reacts (stoichiometry is 2:1:2)

4.9-2x      5.1-x        2x      Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0
2 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
What is the formula of sulfurous acid?<br> H2S<br> HSO4<br> H2SO4<br> H2SO3
Bas_tet [7]

The correct answer is H2SO3 I took the assignment already


6 0
3 years ago
What problem can fossil fuel create for life on earth?
Darya [45]
Pollution and rising temperatures
4 0
3 years ago
Can someone help please !!!!
Naya [18.7K]

Answer:

x = 231/90 Th

Explanation:

3 0
3 years ago
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