<span>The correct answer is b. Radon. Oxygen, Hyrdrogen, and Boron, are not radioactivel ike Radon and it regenerates quickly meaning that even though it has a short half-life period, it stays for a long time once released. It also has no taste so it's difficult to notice without proper gear because you can't feel it.</span>
Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
Answer:

Explanation:
We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.
We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

Flip the ratio so the units of atoms of fluorine cancel each other out.


Condense into 1 fraction.

Divide.

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>
Answer:
I think the answer would be option d.
hope it helps.