Question #5991c

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Answer 1 · 2017-07-23T06:22:01

a. $"Number of molecules" = "Avogradro's contant"*"Number of moles"$

$0.45(6.02*10^(23))=2.709*10^(23)~~2.71*10^(23)$

b. $"Number of atoms" = "Number of molecules"*"Number of atoms in one molecule"$

$NH_4Cl$ = 6 atoms

$6(2.709*10^(23))=1.6254*10^(24)~~1.63*10^(24)$

c. $"Number of atoms in 1 mole"="Avogadro's constant"*"Number of atoms in one molecule"$

$6(6.02*10^(23))=3.612*10^(24)~~3.61*10^(24)$

d. $"Number of atoms of one element"="Number of molecules"*"Number of atoms of one element in one molecule"$

$4(2.709*10^(23))=1.0836*10^(24)~~1.08*10^(24)$

d. $n(NH_4Cl) = 0.45mol$
$n(H_4)$ in $NH_4Cl = 0.45*4/53.5=0.0336448598~~3.36*10^(-2)mol$