Consider the reaction bf3 + nh3 → f3b―nh3
what changes in hybridization (if any) of the b atom are a result of this reaction?
a. before the reaction boron is sp2 hybridized, and after the reaction it is sp3d hybridized.
b. before the reaction boron is sp3 hybridized, and after the reaction it is sp3 hybridized.
c. before the reaction boron is sp2 hybridized, and after the reaction it is sp3 hybridized.
what changes in hybridization (if any) of the n atom are a result of this reaction?
a. before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp2 hybridized.
b. before the reaction nitrogen is sp3 hybridized, and after the reaction it is sp3 hybridized.
c. before the reaction nitrogen is sp3d hybridized, and after the reaction it is sp3 hybridized.

The correct answer is given below

Explanation:

B before the reaction boron is sp2 hybridized and after the reaction boron is it is sp3 hybridized.

    Before the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds between them one is s and rest of the 2 are p.That"s why before the reaction boron is sp2 hybridized.

   Whereas after the reaction boron is joined to 3 fluorine atoms by 3 covalent bonds and one nitrogen atom by one co ordinate bond.That"s why after reaction boron is sp3 hybridized.

B  before the reaction nitrogen is sp3 hybridized and after the reaction it is sp3 hybridized.

  Before the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and contain one lone pair of electron.That"s why before the reaction Nitrogen atom is sp3 hybridized.

 whereas after the reaction Nitrogen is joined to 3 hydrogen atoms by 3 covalent bonds and with one boron atom by a co ordinate bond.That"s why after reaction Nitrogen atom is sp3 hybridized.

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml