Respuesta :
Answer:
Likely [tex]\rm In[/tex] (indium.)
Explanation:
Number of atoms: [tex]N = 2.241 \times 10^{21}[/tex].
Dividing, [tex]N[/tex], the number of atoms by the Avogadro constant, [tex]N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}[/tex], would give the number of moles of atoms in this sample:
[tex]\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}[/tex].
The mass of that many atom is [tex]m = 0.4272\; \rm g[/tex]. Estimate the average mass of one mole of atoms in this sample:
[tex]\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The average mass of one mole of atoms of an element ([tex]114.82\; \rm g \cdot mol^{-1}[/tex] in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass [tex]114.82[/tex]. Indium, [tex]\rm In[/tex], is the closest match.
The symbol of the element is In
Stoichiometry
From the question, we are to determine the identity of the element
First, we will determine the number of moles of sample present
Using the formula
[tex]Number \ of\ moles = \frac{Number\ of \ atoms }{Avogadro's\ constant}[/tex]
Number of moles of the sample = [tex]\frac{2.241\times 10^{21} }{6.022\times 10^{23} }[/tex]
Number of moles of the sample = 0.003721355 mole
Now, we will determine the Atomic mass of the sample
From the formula,
[tex]Atomic\ mass = \frac{Mass}{Number\ of\ moles}[/tex]
Therefore,
Atomic mass of the sample = [tex]\frac{0.4272}{0.003721355}[/tex]
Atomic mass of the sample = 114.8 amu
The element that has an atomic mass of 114.8 amu is Indium. The symbol of Indium is In.
Hence, the symbol of the element is In.
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