In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this

Respuesta:

D

Explicación:

Utilizamos la relación de moles para calcular la presión parcial. El número total de moles es 0.2 + 0.2 + 0.1 = 0.5 moles

Ahora, sabemos que la fracción molar del gas argón es 0.2/0.5

La presión parcial se determina así. Para calcular esto, simplemente multiplicamos el número de moles por la presión total.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

D

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

The result is 14.5 g L⁻¹. Here, the problem indicates to reduce the units to one. The existing units are g/L. To achieve a singular unit format, we can move L to the numerator, which can be executed as per the exponent laws; specifically, 1 / aˣ = a⁻ˣ. Thus, we can express 1 / L as L⁻¹. Consequently, the simplified unit remains g L⁻¹. However, remember to leave a space between two different units. This ultimately depicts a unit of density.

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

10 mL of this solution is further diluted to 250 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

5 mL of this solution is diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 500 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

= 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2