This is about 6-12 days after ovulation
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.
Answer:
They all are made up of cells
Answer:
B) 2
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
The rule apply for the addition and subtraction is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
(3.478-2.31) = 1.168 ≅ 1.17 (Rounded to least decimal digit)
(4.428-3.56) = 0.868 ≅ 0.87 (Rounded to least decimal digit)
So,
1.17 * 0.87 = 1.0 (Rounded to least significant)
Answer - two significant digits
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C