Determining the identity of substances is a critical part of chemistry because once the substance's identity is known, we can predict its behavior and understand the scenarios that it is involved in better.
For example, consider an industrial pipe where fouling (scaling) is occurring. If the compounds present in the scales are identified, steps may be taken to prevent and remove the scaling. This is one of many examples where identifying chemical substances is of high importance.
Answer:
3.0 L of NH₃
Solution:
The equation is as follow,
N₂ + 3 H₂ → 2 NH₃
According to equation,
67.2 L (3 mole) H₂ at STP produces = 44.8 L (3 mole) of NH₃
So,
4.50 L of H₂ will produce = X L of NH₃
Solving for X,
X = (4.50 L × 44.8 L) ÷ 67.2 L
X = 3.0 L of NH₃
Answer:
c.
Explanation:
A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.
The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.
The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.
Chadwick, Thompson, Rutherford, Bohr
D. Electron cloud allowed the particles to pass through