Answer: Adenine and guanine are the two purines and cytosine, thymine and uracil are the three pyrimidines. The main difference between purines and pyrimidines is that purines contain a sixmembered nitrogencontaining ring fused to an imidazole ring whereas pyrimidines contain only a sixmembered nitrogencontaining ring. They both are types or categories of nitrogen containing bases present in nuclei acids of DNA and RNA.
Purines are 2 Ring or Carbon Ring, Nitrogen containing bases. That consist of these 2 rings next placed next to each other. These examples include - Adenine and Guanine.
Pyrimidines are 1 or single Ring Nitrogen containing structures. There are 3 nitrogenous bases that are categorized as pyrimidines. Cytosine, Thymine, and Uracil.
Sr3(PO4)2 is definitely Ionic
Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂

Answer:
397 L
Explanation:
Recall the ideal gas law:

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

The left-hand side is simply some constant. Hence, we can write that:

Substitute in known values:

Solving for <em>V</em>₂ yields:

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.
(Assuming 100 L has three significant figures.)
The answer is visible rays. For example, when a rain storm stops, the sun comes out and you are likely to see a rainbow. This is because the sun shines on water molecules in the air and the colors appear. These colors are called visible light. The only ray that a naked Human eye can see