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:
1.91×1021/6.023×10^23
Explanation:
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molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
<u>Answer:</u> The solubility of carbon dioxide at 5.50 atm is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:
Or,
where,
are the initial concentration and partial pressure of carbon dioxide
are the final concentration and partial pressure of carbon dioxide
We are given:
Putting values in above equation, we get:
Hence, the solubility of carbon dioxide at 5.50 atm is
the modern atomic model shows that electrons are located in a predicted area but cannot be identified in a specific point
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