Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
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Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
Steven Ginsberg named managing editor of The Washington Post, rounding out senior management team
They are alike bc they both have 13 protons and neutrons
Answer:
0.960 m
Explanation:
Given data
- Mass of the solute: 27.9 g
- Molar mass of the solute: 233.2 g/mol
- Mass of the solvent: 125.0 g = 0.1250 kg
First, we will calculate the moles of solute.
27.9 g × (1 mol/233.2 g) = 0.120 mol
The molality of the compound is:
m = moles of solute / kilograms of solvent
m = 0.120 mol / 0.1250 kg
m = 0.960 m
