The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6
Answer:
Compounds whose real bond angle are the same as ideal bond angle;
SF6, BF3, CH4, PCI5
Compounds whose real bond angles differ from ideal bond angles;
H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5
Explanation:
According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.
The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.
For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
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The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.
Answer:
1. stability factor
2. steric hindrance factor
Explanation:
stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.
ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
hope it helps...
The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
Learn more about Groups in the periodic table, here:
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If you combine 24.2 g of a solute that has a molar mass of 24.2 g/mol with 100.0 g of a solvent, what is the molality of the resulting solution
Answer: 10 moles/kg.
Explanation:
Given, Mass of solute = 24.2 g
Molar mass of solute = 24.2 g/mol
[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]
Mass of solvent = 100.0g = 0.1 kg [1 g=0.001 kg]
[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]
Hence, the molality of the resulting solution is 10 moles/kg.
Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis