How many $neutrons$ $"neutrons"$ , massive, neutrally charged nuclear particles, are present in a $2 \cdot m o l$ $2*mol$ quantity of water?

Question

How many $neutrons$ $"neutrons"$ , massive, neutrally charged nuclear particles, are present in a $2 \cdot m o l$ $2*mol$ quantity of water?

1 Answer

Impact of this question

1243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T07:47:48

Well, there are $2 \cdot m o l$ $2*mol$ of water.........and approx. $96 \times 10^{23}$ $96xx10^23$ individual neutrons.

Explanation:

And in each water molecule, i.e. $^{1} H_{2}^{16} O$ $""^1H_2""^16O$ , where we treat the water molecule as AN ISOTOPOMER than contains $8 neutrons$ $"8 neutrons"$ , clearly there are $8 \cdot m o l$ $8*mol$ of neutrons.

And a mole, $Avogadro's Number$ $"Avogadro's Number"$ , specifies $N_{A}$ $N_A$ , $6.022 \times 10^{23}$ $6.022xx10^23$ individual items of stuff. So we have approx. $16 \times N_{A} ≅ 96 \times 10^{23}$ $16xxN_A~=96xx10^23$ individual neutrons.

I really can use mental arithmetic here; the numbers are so impossibly large that $near enuff is good enuff$ $"near enuff is good enuff"$ .

Science

Math

Humanities

... and beyond

How many $neutrons$ $"neutrons"$ , massive, neutrally charged nuclear particles, are present in a $2 \cdot m o l$ $2*mol$ quantity of water?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How many neutrons"neutrons", massive, neutrally charged nuclear particles, are present in a 2⋅mol2*mol quantity of water?

1 Answer

Explanation:

Related questions

Impact of this question

How many $neutrons$ $"neutrons"$ , massive, neutrally charged nuclear particles, are present in a $2 \cdot m o l$ $2*mol$ quantity of water?