QUESTION IMAGE
Question
question 11 blank 1 mols of mg are equal to 2.58x10^26 atoms of mg. blank 1 add your answer question 12 blank 1 x 10 ^ blank 2 atoms are found in a sample contains 2.54 grams of c6h12o6. blank 1 add your answer blank 2 add your answer
Step1: Recall Avogadro's number
Avogadro's number ($N_A$) is $6.022\times 10^{23}$ atoms/mol.
Step2: Calculate moles of Mg for Question 11
$n=\frac{N}{N_A}$, where $N = 2.58\times 10^{26}$ atoms and $N_A=6.022\times 10^{23}$ atoms/mol. So $n=\frac{2.58\times 10^{26}}{6.022\times 10^{23}}$.
$n = \frac{2.58}{6.022}\times10^{26 - 23}\approx0.428\times10^{3}=428$ mol.
Step3: Calculate molar - mass of $C_6H_{12}O_6$ for Question 12
The molar - mass of $C_6H_{12}O_6$: $M=(6\times12.01 + 12\times1.01+6\times16.00)$ g/mol $=(72.06 + 12.12 + 96.00)$ g/mol $ = 180.18$ g/mol.
Step4: Calculate moles of $C_6H_{12}O_6$
$n=\frac{m}{M}$, where $m = 2.54$ g and $M = 180.18$ g/mol. So $n=\frac{2.54}{180.18}$ mol $\approx0.0141$ mol.
Step5: Calculate number of molecules of $C_6H_{12}O_6$
$N=n\times N_A$, so $N = 0.0141\times6.022\times 10^{23}$ molecules $=0.0141\times6.022\times10^{23}$ molecules $\approx8.49\times 10^{21}$ molecules.
Since each $C_6H_{12}O_6$ molecule has $(6 + 12+6)=24$ atoms.
The number of atoms $N_{atoms}=N\times24=(8.49\times 10^{21})\times24 = 203.76\times 10^{21}=2.04\times 10^{23}$ atoms.
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Question 11: 428
Question 12:
Blank 1: 2.04
Blank 2: 23