In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on th

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:

Explanation:

The oxidation state corresponds to the charge of each atomic ion. An increase indicates oxidation of the element while a decrease reflects reduction of the element.

2AgCl+Zn⟶2Ag+ZnCl2

Zinc undergoes oxidation, while Ag experiences reduction.

Ag⁺ changes to Ag (oxidation state decreases), thus Ag is reduced.

Zn alters to Zn⁺² (oxidation state increases), hence Zn is oxidized.

4NH₃+3O₂⟶2N₂+6H₂O

The oxidation state of nitrogen in ammonia is -3

whereas it is zero in elemental nitrogen.

An increase in the oxidation state indicates nitrogen is oxidized.

The oxidation state of oxygen is zero when in molecular oxygen and -2 when in water. Therefore, the oxidation state decreases, indicating oxidation is reduced.

Fe₂O₃+2Al⟶Al₂O₃+2Fe

The oxidation state of Fe in Fe₂O₃ is +3, switching to zero in Fe, so iron is reduced.

Aluminum's oxidation state is zero in Al, rising to +3 in Al₂O₃, indicating it is oxidized.

A skeletal formula is usually used for organic compounds

Density is calculated as mass divided by volume.
  Step one:
Convert m³ to ml.
1 m³ = 1,000,000 ml
0.250 m³ x 1,000,000 = 250,000 ml
  Step two: Convert mg to g.
1 mg = 0.001 g, hence 4.25 x 10^8 mg equals 0.459 g.
Consequently, the density comes out to be 0.459 g/250,000 = 1.836 x 10^-6 g/ml.

1. Stars originate within clouds of gas and dust referred to as nebulas. 2. These clouds are drawn together by gravitational forces. 3. Once sufficient heat and pressure accumulate, nuclear fusion commences, representing the birth of a star.