In the compound iodine heptafluoride: (hints: write out the molecular formula of this compound before answering the question. Also be sure you clearly understand the concepts of charge, oxidation numbers, how to determine charge and oxidation numbers, and - most important of all! - the similarities and the differences between charges and oxidation numbers)
<u>Each fluorine atom has a charge of 1</u>
<h3>What is
iodine heptafluoride?</h3>
The interhalogen compound iodine heptafluoride, often known as iodine(VII) fluoride or iodine fluoride, has the chemical formula IF7. As anticipated by VSEPR theory, it exhibits a unique pentagonal bipyramidal structure. The molecule is capable of undergoing the Bartell process, a pseudorotational rearrangement that is similar to the Berry mechanism but for a heptacoordinated system. It produces colorless crystals that melt at 4.5 °C and have a very narrow liquid range with a boiling point of 4.77 °C. The dense mist has an unpleasant, musty smell. The molecule is symmetrical with D5h. suggestion
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Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
