Answer:
=> The colour of this stone is usually a pale greenish blue, owing to the presence of iron impurities. Stones that are treated with heat look more blue than green. On the Mohs scale of hardness, aquamarine ranges between 7.5 and 8 making it a relatively hard gemstone.
=> The best way to identify a real aquamarine stone is by looking at its colour. In its natural form, they have a pale blue colour, which is similar to seawater. They may have a slight green or yellow tint as well. Naturally occurring gems have excellent clarity and transparency.
=> The hardness of the stone is another feature you can use to identify the stone. Aquamarine stones are hard and they don’t get scratches easily. However, they can easily scratch glass and other such surfaces. So, if you find visible scratches on the stone, rethink your decision to buy it.
=> Most faceted aquamarine stones are clean to the eye and clear of any inclusions. However, translucent and opaque aquamarine is also available. These are usually fashioned into cabochons or beads. In some cases, inclusions may appear as parallel tubes. Such stones can be crafted to show a cat’s eye. Stones with cat’s eye and star effect are rare and highly priced.
The principle quantum number "n" represents the relative overall energy of each orbital, and the energy of each orbital increases as the distance from the nucleus increases. The sets of orbitals with the same "n" value are often referred to as electron shells or energy levels.
Answer:
C
Explanation:
The substance would be classified as neutral, so C.
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
I think the answer is gravity
