Answer:
Neutrons are all identical to each other, just as protons are. Atoms of a particular element must have the same number of protons but can have different numbers of neutrons.
Explanation:
Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).
Answer:
D. 0.75 grams
Explanation:
The data given on the iridium 182 are;
The half life of the iridium 182,
= 15 years
The mass of the sample of iridium, N₀ = 3 grams
The amount left, N(t) after two half lives is given as follows;

For two half lives, t = 2 × 
∴ t = 2 × 15 = 30


∴ The amount left, N(t) = 0.75 grams
<span>your answer is Ca3</span>(PO4)2<span>, </span>
Answer:
2.25×10¯³ mm.
Explanation:
From the question given above, we obtained the following information:
Diameter in micrometer = 2.25 μm
Diameter in millimetre (mm) =?
Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m
2.25 μm = 2.25×10¯⁶ m
Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:
1 m = 1000 mm
Therefore,
2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm
2.25×10¯⁶ m = 2.25×10¯³ mm
Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.