Question #5991c

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Question #5991c

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Answer 1 · 2017-07-23T06:22:01

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

$0.45 (6.02 \cdot 10^{23}) = 2.709 \cdot 10^{23} \approx 2.71 \cdot 10^{23}$ $0.45(6.02*10^(23))=2.709*10^(23)~~2.71*10^(23)$

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

$N H_{4} C l$ $NH_4Cl$ = 6 atoms

$6 (2.709 \cdot 10^{23}) = 1.6254 \cdot 10^{24} \approx 1.63 \cdot 10^{24}$ $6(2.709*10^(23))=1.6254*10^(24)~~1.63*10^(24)$

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

$6 (6.02 \cdot 10^{23}) = 3.612 \cdot 10^{24} \approx 3.61 \cdot 10^{24}$ $6(6.02*10^(23))=3.612*10^(24)~~3.61*10^(24)$

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

$4 (2.709 \cdot 10^{23}) = 1.0836 \cdot 10^{24} \approx 1.08 \cdot 10^{24}$ $4(2.709*10^(23))=1.0836*10^(24)~~1.08*10^(24)$

d. $n (N H_{4} C l) = 0.45 m o l$ $n(NH_4Cl) = 0.45mol$
$n (H_{4})$ $n(H_4)$ in $N H_{4} C l = 0.45 \cdot \frac{4}{53.5} = 0.0336448598 \approx 3.36 \cdot 10^{- 2} m o l$ $NH_4Cl = 0.45*4/53.5=0.0336448598~~3.36*10^(-2)mol$

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Question #5991c

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