Answer: The Lattice energy is the energy required to separate an ionic solid into its component gaseous ions <em>or</em>
It is the energy released when gaseous ions combine to form an ionic solid.
Explanation:
The lattice energy depends on the ionization energies and electron affinities of atoms involved in the formation of the compound. The ionization energies and electron affinities also depends on the ionic radius and charges of the ions involved. As the ionic radius for cations <em>increases</em> down the groups, ionization energy <em>decreases</em>, whereas, as ionic radii <em>decreases</em> across the periods , ionization energy <em>increases</em>. The trend observed for anions is that as ionic radii <em>increase </em>down the groups, electron affinity <em>decreases. </em>Across the period, as ionic radii <em>increases</em> electron affinity <em>increases</em>. Also, as the charge on the ion <em>increases,</em> it leads to an <em>increase</em> in energy requirement/content.
Therefore, for compounds formed from cations and anions in the same period, the highest charged cation and anion will have the highest lattice energy. For example, among the following compounds: Al2O3 (aluminium oxide), AlCl3 (aluminium chloride), MgO, MgCl2 (magnesium chloride), NaCl, Na2O (sodium oxide); Al2O3(aluminium oxide) will have the highest lattice energy, thus will be hardest to break apart because its ions have the highest charge.
Answer:
26.981539 u
Explanation:
One mole of Al atoms has a mass in grams that is numerically equivalent to the atomic mass of aluminum. The periodic table shows that the atomic mass (rounded to two decimal points) of Al is 26.98, so 1 mol of Al atoms has a mass of 26.98 g.
Answer:
The two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place
Explanation:
A complex is made up a central metal atom or ion and ligands. Ligands are lewis bases and they possess lone pairs of electrons. A complex is formed when electrons are donated from ligand species to metals.
However, if the ligand has a negative charge at a particular location and we try to put electrons from the metal near the electrons from the ligand, the two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place.
Answer:
Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)
m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K
Explanation:
Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)
Q= 15750 + 83425 + 105000 + 564000 + 23400
Q= 791575 J
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
To view more about rational reaction, refer to:
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