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Atomic Theory/Nomenclature

Questions:
1) A reaction of one liter of chlorine gas with 3 liters of fluorine gas yields 2 liters of a gaseous product. All gas volumes are at the same temperature and pressure. What is the formula of the gaseous product?

2) Hydrazine, ammonia, and hydrogen azide all contain only nitrogen and hydrogen. The mass of hydrogen that combines with 1.00g of nitrogen for each compound is 1.44*10^-1g, 2.16*10^-1g and 2.40*10^-2g, respectively. Show how these data illustrate the law of multiple proportions?

3) List the noble gas elements. Which of the noble gases has only radioactive isotopes? Which lanthanide element and which transition element have only radioactive isotopes?

4) Name these compounds.
a) Hgsub2O
b) Rbsub2O
c) CaS
d) AlIsub3
e) CsF
f) Lisub3N
g) Agsub2S
h) MnOsub2
i) TiOsub2
j) Srsub3Psub2
k) BaSOsub4
l) NaNOsub2
m) KMnOsub4
n) Ksub2Crsub2Osub7
o) HCsub2Hsub2Osub2
p) NHsub4NOsub2
q) Cosub2Ssub3
r) ICl

5) Write the formula for the compounds. Side note: What is the best way to learn how to name these things?
a) sulfur difluoride
b) sulfur hexafluoride
c) sodium dihydrogen phosphate
d) lithium nitride
e) chromium III carbonate
f) tin II fluoride

Solution Preview

1) You need to remember that at the same temperature and pressure, the volume of a gas is proportional to the number of molecules (or atoms for noble gases) which are in the gas. Chlorine and fluorine are both diatomic, so the ratio of their volumes tells you the ratio of the atoms in the product, and hence the formula.

2) Nitrogen (N2) has a molar mass of 28 g/mol, so 1.00 g is 3.57 x 10-2 mol.
From the molar mass of hydrogen you can similarly calculate the number of moles in 1.44*10^-1g, 2.16*10^-1g and 2.40*10^-2g. Comparing the number of moles of nitrogen to the number of moles of hydrogen gives a simple ratio in each case: the empirical formula of each compound. The law of multiple proportions is ...

Solution Summary

This solution is comprised of a detailed explanation which illustrates how to complete each part of these chemistry based questions. Additionally, tips are provided to help one further understand the concepts. This solution serves as a great learning tool to utilize in order to answer future questions of this nature.

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