Hello Gary My Man!
Well, as you can clearly see
<span>The atomic number of an element is basically the number of protons it has. So yes, for every element this is different. Now, the mass number of an element as known, is the number of protons+the number of neutrons. So theoretically as we can see, this number should be a whole number, but since there are different isotopes (atoms of the same element with different numbers of neutrons) of each element, most periodic tables take account of that, so they often include decimals as seen.
So in Short, ALL</span> the atoms of a particular element have the SAME EXACT atomic number<span> (</span>number<span> of protons of course). The </span>atoms of different elements have very different numbers of protons. And of course, the MASS number of an atom is the TOTAL number as known, of protons and of course, the neutrons it contains in it.
I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)
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-TheOneAboveAll :D
Democritus was the first person to theorize the existence of atoms.
<em>Let </em><em>the </em><em>mass </em><em>be </em><em>X </em><em>g</em>
<em>percentage </em><em>=</em><em> </em><em>X/</em><em> </em><em>6.</em><em>5</em><em>0</em><em> </em><em>*</em><em> </em><em>100 </em><em>=</em><em>2.</em><em>2</em><em>%</em>
<em>X=</em><em> </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
<em>The </em><em>mass </em><em>is </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
