Answer:
Most of the elements in the periodic table are <u><em>metals</em></u>.
Hope that helps. x
Arrhenius Theory: according to Arrhenius, acid is one that can donate proton in an aqueous solution, while base is one that can donate hydroxide ion in an aqueous solution.
Bronsted-Lowry Theory: according to Bronsted Lowry, acid is one that can donate protons while base is one that can accept a proton.
1. In first, only C. NH3 can't give hydroxide ion, but can accept a proton so it is a Bronsted-Lowry Base but not an Arrhenius base.
2.In second, as the definition suggested, bronsted base is one that can accept protons and acid is one that can loose protons. so answer is D. Acids lose H+ and bases gain H+.
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Answer:
ΔH = ΔH₁ + ΔH₂ - ΔH₃
Explanation:
Given that:
1. A → 2B
2. B → C + D
3. E → 2D
Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.
To estimate the ΔH for the process A → 2C + E
We multiply 2 with equation 2 where (B → C + D)
2B → 2C + 2D ⇒ 2ΔH₂
Also, let's switch equation (3), such that we have,
2D → E -ΔH₃
The summation of all the equation result into :
A → 2C + E
where; ΔH = ΔH₁ + ΔH₂ - ΔH₃
The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.
There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.
B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.
The basis of ionic compounds are ions and the basis of polar compounds are dipoles.
The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.
Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.
When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.
Then, the answer is dispersion interaction.
