Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
Answer:
the volume of a give gas simple is directly propotional assolute temperature at constant pressure .the volume of a gavi. amount of gass is inversely propotional ot Its pressure when temperature is help constant
Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
Answer:
Mass of heptane = 102g
Vapor pressure of heptane = 454mmHg
Molar mass of heptane = 100.21
No of mole of heptane = mass/molar mass = 102/100.21
No of mole of heptane = 1.0179
Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane
Partial pressure of heptane = 1.0179*454mmHg
Partial pressure of heptane = 462.1096 = 462mmHg
the partial pressure of heptane vapor above this solution = 462mmHg
The complete balanced chemical equation for this is:
<span>3KOH + H3PO4
--> K3PO4 + 3H2O</span>
First we calculate the number of moles of H3PO4:
moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol
From stoichiometry, 3 moles of KOH is required for every
mole of H3PO4, therefore:
moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole
H3PO4) = 0.0468 mol
Calculating for volume given molarity of 0.350 M KOH:
Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7
mL
Answer:
<span>133.7 mL KOH</span>
