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4 years ago

What is the empirical formula for Hg2(NO3)2

1 answer:

8 0

Answer: HgNO₃


Explanation:


The empirical formula is the formula that shows the ratio of the atoms in its simplest form, this is using the smallest whole numbers.


The empirical formula may or may not be the same molecular formula.


In this case you are given the molecular formula Hg₂(NO₃)₂. Since, the ratio of the atoms of Hg, N, and O is 2: 2: 6, respectively, the same ratio is expressed if you divide by the greatest common factor (GCF).


The GCF of 2, 2, and 6 is 2. So, the ratios can be simplified to 1:1:3, meaning 1 mol of Hg, 1 mol of N, and 3 mol of O or HgNO₃.

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When 20.4g of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride

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Answer: True

Explanation:

$\mathrm{The\ balanced \ chemical \ equation\ is\ :}\\$2 \mathrm{Na}{(s)}+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(s)$

For Na

$\\$Given mass $=20.4 \mathrm{~g}$ \ \ Molecular mass $=23 \mathrm{~g}\ \ $Number of moles $=\frac{\text { Given mass }}{\text { Molecular mass }}$$$\begin{aligned}&=\frac{20.4}{23} \\&=0.887 \text { moles }\end{aligned}$$Stoichiometric ratio of $\mathrm{Na}$ is same as that of $\mathrm{NaCl}$ i.e., $1: 1$.$\therefore$ Moles of $\mathrm{NaCl}=0.887$ motes$

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$\\\begin{aligned}\text { Molecular mass } &=23+35.5 \\&=58.5 \mathrm{~g}\end{aligned}$

$\begin{aligned}\\& $\therefore$ Mass of $\mathrm{NaCl}$ produced will be=\text { No.of moles } \times \text { Molecular mass } \\&=0.887 \times 58.5 \\&=51.9 \\&\simeq 52 \mathrm{~g} \quad\end{aligned}$$$

Therefore, it is true that when 20.4g of sodium metal are mixed with chlorine gas, 52.0 g of sodium chloride is produced

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