# In 6.72 g of #"N"_2"H"_4#, how many...?

##
(a) how many #"N"_2"H"_4# molecules are present?

(b) how many #"N"# atoms are present?

(c) how many protons are present

(a) how many

(b) how many

(c) how many protons are present

Here's what I got.

Your tool of choice for this problem will be Avogadro's number, which tells you how many molecules of a substance you get in one mole of that substance.

#color(blue)(|bar(ul(color(white)(a/a)"one mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) -># Avogadro's number

So, you know that one mole of *hydrazine*, *how many moles* you have in that

To figure that out, use hydrazine's molar mass, which tells you the mass of one mole of hydrazine molecules. Since hydrazine has a molar mass of

This means that your sample contains

#6.72color(red)(cancel(color(black)("g"))) * overbrace(("1 mole N"_2"H"_4)/(32.045color(red)(cancel(color(black)("g")))))^(color(brown)("molar mass of N"_2"H"_4)) = "0.2097 moles N"_2"H"_4#

Now that you *how many moles* you have, use Avogadro's number to find how many *molecules* you have

#0.2097color(red)(cancel(color(black)("moles"))) * overbrace((6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))))^(color(purple)("Avogadro's number")) = color(green)(|bar(ul(color(white)(a/a)1.26 * 10^(23)"molecules"color(white)(a/a)|)))#

Now, for part (b) you must find the number of nitrogen atoms,

two atoms of nitrogen,#2 xx "N"# four atoms of hydrogen,#4 xx "H"#

So, if you get

#2# atoms of nitrogen for every molecule of hydrazine, you can say that the total number of nitrogen atoms will be

#1.26 * 10^(23)color(red)(cancel(color(black)("molecules N"_2"H"_4))) * "2 atoms of N"/(1color(red)(cancel(color(black)("molecule N"_2"H"_4)))) = color(green)(|bar(ul(color(white)(a/a)2.52 * 10^(23)"atoms of N"color(white)(a/a)|)))# Finally, for part (c) you must find the total number of protons present in the sample. As you know, the number of protons present in the nucleus of a given atom is equal to that atom's atomic number.

A quick look in the periodic table will reveal that nitrogen has an atomic number equal to

#7# , which means that every atom of nitrogen contains#7# protons.Hydrogen has an atomic number equal to

#1# , which means that every atom of hydrogen contains#1# proton in its nucleus.Now, you already know how many atoms of nitrogen you have in the sample. Use the same approach to find the number of atoms of hydrogen

#1.26 * 10^(23)color(red)(cancel(color(black)("molecules N"_2"H"_4))) * "4 atoms of H"/(1color(red)(cancel(color(black)("molecule N"_2"H"_4)))) = 5.04 * 10^(23)"atoms of H"# This means that the total number of protons present in the sample will be

#"no. of protons" = overbrace(7 xx 2.52 * 10^(23))^(color(blue)("coming from nitrogen")) + overbrace(1 xx 5.04 * 10^(23))^(color(red)("coming from hydrogen"))#

#"no. of protons" = color(green)(|bar(ul(color(white)(a/a)2.27 * 10^(24)"protons"color(white)(a/a)|)))# The answers are rounded to three sig figs, the number of sig figs you have for the mass of hydrazine.

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There are about 0.4 moles of N2H4 in 6.72 g of N2H4.

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*Answer from HIX Tutor*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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