Answer:
209.8 kilo Joules heat is required to vaporize 250 g of ethanol at 78.5 °C.
Explanation:
Mass of an ethanol = 250 g
Molar mass of ethanol = 46 g/mol
Moles of an ethanol = 
Enthalpy of vaporization of ethanol = 
Heat required to vaporize 250 g of ethanol at 78.5 °C : Q
Q = 209.8 kilo Joules
Answer:
Graphitic carbon nitride (g-C3N4) is a rising two-dimensional material possessing intrinsic semiconducting property with unique geometric configuration featuring superimposed heterocyclic sp2 carbon and nitrogen network, nonplanar layer chain structure, and alternating buckling. The inherent porous structure of heptazine-based g-C3N4 features electron-rich sp2 nitrogen, which can be exploited as a stable transition metal coordination site. Multiple metal-functionalized g-C3N4 systems have been reported for versatile applications, but local coordination as well as its electronic structure variation upon incoming metal species is not well understood. Here we present detailed bond coordination of divalent iron (Fe2+) through micropore sites of graphitic carbon nitride and provide both experimental and computational evidence supporting the aforementioned proposition.
Answer:
On the Moh's scale of hardness, aluminum oxide is positioned just below to diamond due to which it is considered as one of the hardest known compounds. This also shows that the compound exhibit an enormous amount of lattice energy, as to transform the oxide into its constituent ions, the energy is required to overcome.
Based on the chemical formula of the compound, that is, Al2O3, it is shown that the ions of Al3+ and O2- are kept close due to the activity of the strong electrostatic ionic bonds. The electrostatic forces and the ionic bonding between the ions are extremely robust due to the presence of the ions high charge density. Therefore, to dissociate the bonds, an enormous amount of energy is needed, and at the same time, a high amount of lattice energy is present.
Answer:
NH3 Ammonia
Explanation:
Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.
Answer:
he predicted the properties from known elements above and belws the unknown in the same group
Explanation:
What allowed Mendeleev to make predictions of undiscovered elements
He realized that an element on this table with one known element above it and one known element below it had to have properties between the two known elements
How did Mendeleev predict gallium and germanium?
Based on gaps in the periodic table Mendeleev deduced that in these gaps belonged elements yet to be discovered. Based on other elements below and above in the same group he predicted the existence of eka-aluminum, eka-boron, and eka-silicon, later to be named gallium (Ga), scandium (Sc), and germanium (Ge).
