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>.
Answer:
0.052mL
Explanation:
1mole of a gas occupy 22.4L.
Therefore, 1 mole of CO2 will also occupy 22.4L.
If 1mole of CO2 occupies 22.4L,
Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L
coverting this volume to mL, we simply divide by 1000 as shown below:
51.52/1000 = 0.05152mL = 0.052mL
the formula of the heavy water is D2O
The answer is D. hope I was right
Answer:
B: Adding water, then adding solute
Explanation:
This is because, say you have a solution with a certain concentration.
If you add more water, it will become more diluted (less concentrated)
If you add more solute, it will become more concentrated.
Therefore if you add water and solute, it could cancel out, and the concentration would remain the same.
Hope this helps! Let me know if you have any questions/ would like anything further explained :)
